Concise Mathematics Class 10 ICSE Solutions for Chapter 5 - Quadratic Equations

Exercise 5(A)

1. Find which of the following equations are quadratic:

(i) (3x - 1)² = 5(x + 8)

Ans:  Solving the given equation

(3x - 1)² = 5(x + 8)

⇒ (9x² - 6x + 1) = 5x + 40

⇒ 9x² - 11x - 39 =0

which is of the form ax2 + bx + c = 0.

∴ Given equation is a quadratic equation.

(ii) 5x² - 8x = -3(7 - 2x)

Ans:  Solving the given equation

5x² - 8x = -3(7 - 2x)

⇒ 5x² - 8x = 6x - 21

⇒ 5x² - 14x + 21 =0 

which is of the form ax2 + bx + c = 0.

∴ Given equation is a quadratic equation.

(iii) (x - 4)(3x + 1) = (3x - 1)(x +2)

Ans:  Solving the given equation 

(x - 4)(3x + 1) = (3x - 1)(x +2)

⇒ 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2

⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.

∴ Given equation is not a quadratic equation. 

(iv) x² + 5x - 5 = (x - 3)2

Ans:  Solving the given equation 

x² + 5x - 5 = (x - 3)²

⇒ x² + 5x - 5 = x² - 6x + 9

⇒ 11x - 14 =0; which is not of the form ax2 + bx + c = 0.

∴ Given equation is not a quadratic equation. 

(v) 7x3 - 2x² + 10 = (2x - 5)²

Ans:  Solving the given equation 

7x³ - 2x² + 10 = (2x - 5)²

⇒ 7x³ - 2x² + 10 = 4x² - 20x + 25

⇒ 7x³ - 6x² + 20x - 15 = 0

 which is not of the form ax2 + bx + c = 0.

∴ Given equation is not a quadratic equation.

(vi) (x - 1)2 + (x + 2)2 + 3(x +1) = 0

Ans:  Solving the given equation 

(x - 1)² + (x + 2)² + 3(x +1) = 0

⇒ x² - 2x + 1 + x² + 4x + 4 + 3x + 3 = 0

⇒ 2x² + 5x + 8 = 0 

which is of the form ax² + bx + c = 0.

∴ Given equation is a quadratic equation.

2. (i) Is x = 5 a solution of the quadratic equation x² - 2x - 15 = 0?

Ans:  P(x)=x² - 2x - 15 = 0

For x = 5 to be the solution of the given quadratic equation it should satisfy the equation. i.e P(5) = 0

So, substituting x = 5 in the given equation, we get

P(5) = (5)² - 2(5) - 15= 25 - 10 - 15 = 0

Hence, x = 5 is a solution of the quadratic equation x² - 2x - 15 = 0. 

(ii) Is x = -3 a solution of the quadratic equation 2x2 - 7x + 9 = 0?

Ans:  2x² - 7x + 9 = 0

For x = -3 to be the solution of the given quadratic equation it should satisfy the equation. i.e P(-3)=0

So, substituting x = -3 in the given equation, we get

P(-3)=2(-3)2 - 7(-3) + 9 = 18 + 21 + 9  = 48 ≠ 0

Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0. 

3. If $\sqrt {\frac{2}{3}} $ is a solution of equation 3x² + mx + 2 = 0, find the value of m.

Ans:  For x =  $\sqrt {\frac{2}{3}} $ to be solution of the given quadratic equation it should satisfy the equation

So, substituting x =$\sqrt {\frac{2}{3}} $ in the given equation, we get

 3 ${\left( {\sqrt {\frac{2}{3}} } \right)^2}$ + m $\left( {\sqrt {\frac{2}{3}} } \right)$ + 2 = 0

⇒3 $\left( {\sqrt {\frac{2}{3}} } \right)$ + m $\left( {\sqrt {\frac{2}{3}} } \right)$ + 2 = 0

⇒-4 $\left( {\sqrt {\frac{3}{2}} } \right)$ = m

⇒-2($\sqrt 6 $)  = m

Hence for m=-2($\sqrt 6 $),   x =  $\sqrt {\frac{2}{3}} $ to is solution of 3x2 + mx + 2 = 0

4. $\frac{2}{3}$ and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.

Ans:  For x =$\frac{2}{3}$ and x = 1 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = $\frac{2}{3}$ and x = 1 in the given equation, we get

m ${\left( {\frac{2}{3}} \right)^2}$ + n $\left( {\frac{2}{3}} \right)$ + 6 = 0

⇒ m $\left( {\frac{4}{9}} \right)$ + n $\left( {\frac{2}{3}} \right)$ + 6 = 0

⇒ 4m + 6n + 54 = 0………. (i)

m($1$)² + n($1$)+ 6 = 0

⇒ m + n +6 = 0………..(2)

Solving equations (1) and (2) simultaneously,

Multiply equation (2) by 6 and subtract with equation (1), 

-2m +18 = 0

⇒ m = 9

Substitute value of m in (2),

n + 9 + 6 = 0

⇒ n = -15

Thus, the value of m is 9 and the value of n is −15

5. If 3 and -3 are the solutions of equation ax² + bx - 9 = 0. Find the values of a and b.

Ans:  For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = 3 and x = -3 in the given equation, we get

a(3)² + b(3) - 9 = 0

⇒ a(9) + b(3) - 9 = 0

⇒ 3a + b - 3 = 0………..(1)

a(-3)² + b(-3) - 9 = 0

⇒ 9a -3b - 9 = 0

⇒ 3a -b -3 = 0……….(2)

Solving equations (1) and (2) simultaneously,

Adding equation (1) and equation (2),

(3a + b - 3)+ (3a -b -3) = 0

⇒ 6a -6 = 0

⇒ a = 1

Substitute value of a in ( 1)

3(1) -b -3 = 0

⇒b = 0

Exercise 5(B)

1. Without solving, comment upon the nature of roots of each of the following equations:

(i) 7x² - 9x +2 =0

Ans:  a =7, b= -9, c=2

∴ Discriminant = b² - 4ac

                      D = (-9)² -4(7)(2)

                      D = 81-56

                      D = 25 > 0

Since D > 0 and 25 is not a perfect square. Therefore, the equation has two real and unequal roots.

(ii) 6x² - 13x +4 =0

Ans:  a = 6 , b = -13, c = 4

∴ Discriminant = b² - 4ac

                    D = (-13)² - 4(6)(4)

                    D = 169 - 56

                    D = 113 > 0

Since D > 0. Therefore, the equation has two real and unequal roots.

(iii) 25x² - 10x +1=0

Ans:  a = 25, b = -10, c = 1

∴ Discriminant = b² - 4ac

                      D = (-10)² - 4(25)(1)

                      D = 100 - 100

                      D = 0

Since D = 0. Therefore, the equation has two real and equal roots.

(iv) x² +2($\sqrt 3 $)x- 9 = 0

Ans:  a = 1, b = 2$\sqrt 3 $, c = -9

∴ Discriminant = b² - 4ac

                      D = (2$\sqrt 3 $)² -4(1)(-9)

                      D = 12+36= 48 >0

Since D > 0 and 25 is not a perfect square. Therefore, the equation has two real and unequal roots.

(v) x² - ax - b² =0

Ans:  a = 1, b = -a, c = -(b)²

∴ Discriminant = b² – 4ac

                     D = (-a)² – 4 (1) (-b)²

                     D = a² + 4b² >0                        

Since D > 0. Therefore, the equation has two real and unequal roots.

(vi) 2x² + 8x + 9 = 0

Ans:   a = 2, b = 8, c = 9

∴ Discriminant = b² - 4ac

                         = (8)² - 4(2)(9)

                         = 64-72

                         = -8 < 0 

Since D <0. Therefore, the equation has no real roots.

2. (i) Find the value of p, if the following quadratic equation has equal roots : 4x² - (p - 2)x + 1 = 0

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac =0

Here, a = 4, b = -(p-2), c = 1

    D = (p-2)² - 4(4)(1) = 0

⇒ (p-2)² -(4)2= 0

⇒ (p-2+4)(p-2-4)= 0

⇒ (p+2)(p-6)= 0

Thus, the values of p are −2 and 6                        

(ii). Find the value of 'p', if the following quadratic equations have equal roots:x² + (p - 3)x + p = 0

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac =0

Here, a = 1, b = (p-3), c=p

D = (p-3)² - 4(1)(p) = 0

⇒ (p)²-6p+9-4p= 0

⇒ (p)²-10p+9= 0

⇒ (p)²-p-9p+9= 0

⇒ p(p-1)-9(p-1)= 0

⇒ (p-1)(p-9)= 0

Thus, the values of p are 1 and 9

3. The equation 3x² - 12x + (n - 5) = 0 has equal roots. Find the value of n.

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac =0

Here, a =3, b = -12, c = n-5

D = (-12)² -4(3)(n-5) = 0

⇒144-12(n-5) = 0

⇒ 144 + 60 -12n= 0

⇒ 204= 12n 

⇒ n= $\frac{{204}}{{12}}$

⇒ n = 17

Thus, the value of n is =17

4. Find the value of m, if the following equation has equal roots: (m - 2)x² - (5+m)x +16 =0

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac =0

Here, a = m-2, b= m + 5, c = 16

D = (m+5)²-4(m-2)(16) = 0

⇒ (m)² +10m+25- 64(m-2)= 0

⇒ m²-54m+153= 0

⇒ (m)²-3m-51m+153= 0

⇒ m(m-3)-51(m-3)= 0

⇒ (m-3)(m-51)= 0

Thus, the values of m are 51 and 3.

 5. Find the value of k for which the equation 3x²- 6x + k = 0 has distinct and real roots.

Ans:  Since the equation has distinct roots. Therefore, the discriminant will be greater than zero.

∴ Discriminant = b² - 4ac > 0

Here, a = 3, b = -6, c = k

D = (-6)² - 4(3)(k) > 0

⇒ 36 -12k > 0

⇒ 36 > 12k

⇒ k < 3

Thus, the value of k is 3 less than 3.

Exercise 5(C)

Solve equations number 1 to number 20, given below using factorization method

1. Solve x²-10x-24 =0

Ans:  x²-10x-24 =0

⇒ x²-12x+2x-24 =0

⇒ x(x-12)+2(x-12) =0

⇒ (x+2)(x-12) =0

Since, x-12 = 0 or x+2 = 0

Then   x =12    or x =-2

2. Solve x²-16 =0

Ans:  x²-16 =0

⇒ x²-(4)² =0

⇒ (x+4)(x-4) =0

Since, x-4 =0 or x+4 = 0

Then x =4 or x =-4

3. Solve 2x²- $\frac{1}{2}$x =0

Ans:  2x² - $\frac{1}{2}$x = 0

⇒ x $\left( {2x - \frac{1}{2}} \right)$ = 0

Since, x=0 or 2x-$\frac{1}{2}$ = 0

Then   x =0    or x = $\frac{1}{4}$

4. Solve x(x-5)=24

Ans:  x²-5x-24 =0

⇒ x²-8x+3x-24 =0

⇒ x(x-8)+3(x-8) =0

⇒ (x+3)(x-8) =0

Since, x-8 =0 or x+3 = 0

Then x =9 or x =-3

5. Solve $\frac{9}{2}$ x = x² + 5 

Ans:  $\frac{9}{2}$ x = x² + 5 

⇒ $9$ x = 2x² + 10 

⇒ 2x² - 9x + 10 = 0

⇒ 2x² - 4x - 5x + 10 = 0

⇒ 2x(x-2) - 5(x-2) =0

⇒ (2x-5)(x-2) =0

Since, x-2 =0 or 2x-5 = 0

Then   x =2 or x =$\frac{5}{2}$

6. Solve $\frac{6}{x}$ = 1+x

Ans:  x²+x =6

⇒ x²+x-6 =0

⇒ x²+3x -2x-6 =0

⇒ x(x+3)-2(x+3) =0

⇒ (x+3)(x-2) =0

Since, x-2 = 0 or x+3 = 0

Then x =2 or x =-3

7. Solve: x =$\frac{{3x + 1}}{{4x}}$

Ans:  4x² = 3x +1

⇒ 4x²-3x-1 = 0

⇒ 4x²-4x+x-1 = 0

⇒ 4x(x-1)+1(x-1) =0

⇒ (4x+1)(x-1) =0

Since, 4x+1 = 0 or x-1 = 0

Then x = -$\frac{1}{4}$ or x =1

8. Solve: x + $\frac{1}{x}$= 2.5

Ans:  x + $\frac{1}{x}$= $\frac{5}{2}$

⇒ x + $\frac{1}{x}$= $\frac{5}{2}$

⇒ 2(x² + 1) = 5x 

⇒ 2(x² + 1) = 5x 

⇒ 2x² -5x+ 2 = 0

⇒ 2x² -4x - x + 2 = 0

⇒ 2x(x-2) -1(x- 2) = 0

⇒ (2x-1)(x-2) = 0

Since, 2x-1 = 0 or x-2 = 0

Then x = $\frac{1}{2}$ or x = 2

9. Solve: (2x-3)² = 49

Ans:  (2x-3)² =49

⇒      4x²-12x+9-49 =0

⇒      4x²-12x-40 =0

⇒      x²-3x-10 =0

⇒      x²-5x+2x-10 =0

⇒      x(x-5) +2(x- 5) = 0

⇒      (x-2)(x-5) =0

Since, x+2 =0 or x-5 = 0

Then x =-2 or x =5

10.Solve 2(x2-6)=3(x-4)

Ans:  2(x²-6)=3(x-4)

⇒      2x²-3x=0 

⇒      x(2x-3) = 0 

Since, x =0 or 2x-3 = 0

Then x =0 or x =$\frac{3}{2}$

11. Solve (x+1)(2x + 8) = (x+7)(x+3)

Ans:  (x+1)(2x + 8) = (x+7)(x+3)

⇒      2x²+8x+2x+8 = x²+7x+3x+21

⇒      x²-13 = 0

⇒      (x+$\sqrt {13} $)(x-$\sqrt {13} $) =0

Since, x-$\sqrt {13} $ =0 or x+$\sqrt {13} $ =0

Then x=$\sqrt {13} $  or x= -$\sqrt {13} $

12. Solve x²-(a+b)x+ab =0

Ans:  x²-(a+b)x+ab =0

⇒      x²-ax-bx+ab =0

⇒      x(x-a)-b(x-a) =0

⇒      (x-a)(x-b) =0

Since, x-a=0 or x-b= 0

Then x =a or x =b

13. Solve (x+3)²-4(x+3)-5 =0

Ans:  (x+3)2-4(x+3)-5 =0

⇒      x²+6x+9-4x-12-5 =0

⇒      x²+2x-8 =0

⇒      x²+4x-2x-8 =0

⇒      x(x+4) -2(x+4) = 0

⇒      (x-2)(x+4) =0

Since, x+4 =0 or x-2 = 0

Then x =-4 or x =2

14. Solve 4(2x-3)²-(2x-3)-14 =0

Ans:  (2x-3)²-(2x-3)-14 =0

Put 2x-3 = y

⇒      4y²-y-14 =0

⇒     4y²-8y+7y-14 =0

⇒      4y(y-2)+7(y-2) =0

⇒      (4y+7)(y-2) =0

Since, 4y+7 =0 or y-2 = 0

 ⇒  4(2x-3)+7 = 0 or (2x-3)-2 = 0

 ⇒   8x =12-7 or 2x =5 

Then x = $\frac{5}{8}$ or x = $\frac{5}{2}$

15. Solve $\frac{{3x - 2}}{{2x - 3}}$= $\frac{{3x - 8}}{{x + 4}}$

Ans:  $\frac{{3x - 2}}{{2x - 3}}$= $\frac{{3x - 8}}{{x + 4}}$

⇒      (3x-2)(x+4)=(3x-8)(2x-3)

⇒    3x²+12x-2x-8 = 6x2-9x-16x+24

⇒    3x²+10x-8 = 6x2-25x+24

⇒     3x²-35x+32 = 0

⇒    3x²-32x-3x+32 = 0

⇒    x(3x-32)-1(3x-32) = 0

⇒    (x-1)(3x-32) = 0

Since, x-1=0 or 3x-32 = 0

Then x =1 or x =$\frac{{32}}{3}$

16. Solve 2x² - 9x + 10 = 0, When

(i) x ∈ N

(ii) x ∈ Q

Ans:  2x²-9x+10 =0

⇒      2x²-4x-5x+10 =0

⇒      2x(x-2)-5(x-2) =0

⇒      (2x-5)(x-2) =0

Since, 2x-5=0 or x-2 = 0

Then x = $\frac{5}{2}$ or x =2

(i) When x ∈ N, x =2

(ii) When x ∈ Q, x = $\frac{5}{2}$ or x =2

17. $\frac{{x - 3}}{{x + 3}}$+$\frac{{x + 3}}{{x - 3}}$=2 $\frac{1}{2}$

Ans:  For  $\frac{{x - 3}}{{x + 3}}$ + $\frac{{x + 3}}{{x - 3}}$ = 2 $\frac{1}{2}$

⇒ $\frac{{\left( {x - 3} \right)\left( {x - 3} \right) + \left( {x + 3} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}$ = $\frac{5}{2}$

⇒ 2[(x-3)²+(x+3)²]=5(x+3)(x-3)

⇒ 2(2x²+18)=5(x²-9)

⇒ x²-81=0

⇒ (x-9)(x+9)=0

Since, x-9=0 or x+9 = 0

⇒ x = 9 or x = -9

18. Solve $\frac{4}{{x + 2}}$ - $\frac{1}{{x + 3}}$ = $\frac{4}{{2x + 1}}$

Ans:  For $\frac{4}{{x + 2}}$-$\frac{1}{{x + 3}}$= $\frac{4}{{2x + 1}}$

⇒$\frac{{4\left( {x + 3} \right) - \left( {x + 2} \right)}}{{\left( {x + 3} \right)\left( {x + 2} \right)}}$= $\frac{4}{{2x + 1}}$

⇒$\frac{{3x + 10}}{{\left( {x + 3} \right)\left( {x + 2} \right)}}$= $\frac{4}{{2x + 1}}$

⇒ (2x+1)(3x+10)=4(x+3)(x+2)

⇒ 6x2+20x+3x+10=4(x2+5x+6)

⇒ 2x²+3x-14=0

⇒ 2x²-4x+7x-14=0

⇒ 2x(x-2)+7(x-2)=0

⇒ (x-2)(2x+7)=0

Since, x-2=0 or 2x+7 = 0

⇒ x=2 or x=-$\frac{7}{2}$

19. Solve $\frac{5}{{x - 2}}$-$\frac{3}{{x + 6}}$= $\frac{4}{x}$

Ans:  For $\frac{5}{{x - 2}}$-$\frac{3}{{x + 6}}$= $\frac{4}{x}$

⇒ $\frac{{5\left( {x + 6} \right) - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 6} \right)}}$= $\frac{4}{x}$

⇒ $\frac{{2x + 36}}{{\left( {x + 6} \right)\left( {x - 2} \right)}}$= $\frac{4}{x}$

⇒ (2x+36)(x)=4(x+6)(x-2)

⇒ 2x²+36x=4(x2+4x-12)

⇒ x²-10x-24=0

⇒ x²-12x+2x-24=0

⇒ x(x-12)+2(x-12)=0

⇒ (x+2)(x-12)=0

Since, x-12=0 or x+2 = 0

⇒ x=12 or x=-2

20. (1+$\frac{1}{{x + 1}}$)(1-$\frac{1}{{x - 1}}$)= $\frac{7}{8}$

Ans:  (1+$\frac{1}{{x + 1}}$)(1-$\frac{1}{{x - 1}}$)= $\frac{7}{8}$

⇒ ($\frac{{x + 1 + 1}}{{x + 1}}$)($\frac{{x - 1 - 1}}{{x - 1}}$)= $\frac{7}{8}$

⇒ ($\frac{{x + 2}}{{x + 1}}$)($\frac{{x - 2}}{{x - 1}}$)= $\frac{7}{8}$

⇒ 8(x+2)(x-2)=7(x+1)(x-1)

⇒ 8(x²-4)=7(x2-1)

⇒ x²-25=0

⇒ (x+5)(x-5)=0

Since, x-5=0 or x+5 = 0

⇒ x=5 or x=-5 

21. Find the quadratic equation, whose solution set is:

(i) {3,5}

Ans:  Since the solution set is {3,5}.

Therefore, the equation for the given solution set can be written as, 

 (x-5)(x-3) =0

⇒ x²-5x-3x+15 = 0

⇒ x²-8x+15 =0

 Thus, the required quadratic equation is x²-8x+15 =0

(ii){-2,3}

Ans:  Since the solution set is {-2,3}.

Therefore, the equation for the given solution set can be written as, 

 (x+2)(x-3) =0

⇒ x²-3x+2x-6 =0

⇒ x²-x-6=0

Thus, the required quadratic equation is x²-x-6=0

22. (i)Solve : $\frac{x}{3}$+$\frac{3}{{6 - x}}$= $\frac{{2\left( {x + 6} \right)}}{{15}}$

Ans:  For $\frac{x}{3}$+$\frac{3}{{6 - x}}$= $\frac{{2\left( {x + 6} \right)}}{{15}}$

⇒$\frac{{x\left( {6 - x} \right) + 9}}{{3\left( {6 - x} \right)}}$= $\frac{{2\left( {x + 6} \right)}}{{15}}$

⇒$\frac{{x\left( {6 - x} \right) + 9}}{{3\left( {6 - x} \right)}}$= $\frac{{2\left( {x + 6} \right)}}{{15}}$

⇒15 [x(6-x) +9] =6(6+x)(6-x)

⇒15(-x²+6x+9)=6(-x2+36)

⇒ x²-10x+9=0

⇒ x²-x-9x+9=0

⇒ x(x-1)-9(x-1)=0

⇒ (x-9)(x-1)=0

Since, x-9=0 or x-1 = 0

⇒ x=9 or x=1                      

(ii) Solve the equation 9x2 + $\frac{3}{4}$x +2 =0

Ans:  For 9x² + $\frac{3}{4}$x +2 =0

⇒36x²+3x+8=0

a=36, b=3, c=8

∴ Discriminant = b² - 4ac

                      D = (3)² -4(36)(8)

                      D = 9-1152

                      D = -1143 < 0

Since D < 0. Therefore, the given equation has no real roots.

23. Find the value of x, if a + 1=0 and x2 + ax - 6 =0.

Ans:  If a+1=0, then a = -1

Put this value in the given equation x2 + ax - 6 =0

          x²-x-6 =0

⇒      x²-2x+3x-6 =0

⇒      x(x-2)+3(x-2) =0

⇒      (x+3)(x-2) =0

Since, x-2 = 0 or x+3 = 0

Then x =2 or x =-3

24. Find the value of x, if a + 7 = 0; b + 10 = 0 and 12x² = ax - b.

Ans:  If a + 7 =0, then a = -7 and b + 10 =0, then b = - 10 

Put these values of a and b in the given equation

          12x²=-7x-(-10)

⇒      12x²+7x-10=0

⇒      12x²-8x+15x-10 =0

⇒      4x(3x-2)+3(3x-2) =0

⇒      (4x+3)(3x-2) =0

Since, 4x+3=0 or 3x-2 = 0

Then x =-$\frac{3}{4}$ or x =$\frac{2}{3}$

25. Use the substitution y= 2x +3 to solve for x, if 4(2x+3)² - (2x+3) - 14 =0.

Ans:  4(2x+3)2 - (2x+3) - 14 =0

         Put 2x+3 = y

⇒      4y²-y-14 =0

⇒     4y²-8y+7y-14 =0

⇒      4y(y-2)+7(y-2) =0

⇒      (4y+7)(y-2) =0

Since, 4y+7 =0 or y-2 = 0

 ⇒ 4(2x+3)+7 =0 or (2x+3)-2 = 0

 ⇒ 8x =-12-7    or   2x =2-3 

Then x =- $\frac{{19}}{8}$ or x =- $\frac{1}{2}$

26. Without solving the quadratic equation 6x² - x – 2 = 0, find whether x = $\frac{2}{3}$ is a solution of this equation or not.

Ans:  Consider the equation, 6x² - x - 2=0, If x = $\frac{2}{3}$ is a root of the equation P(-1) must be 0,

Put x = $\frac{2}{3}$ in P(x)

P($\frac{2}{3}$) =  6($\frac{2}{3}$)2 - $\frac{2}{3}$ - 2

         = 6($\frac{4}{9}$)- $\frac{8}{3}$ 

        = $\frac{8}{3}$- $\frac{8}{3}$=0

Hence, x =  $\frac{2}{3}$ is a solution of the given equation.

27. Determine whether x = -1 is a root of the equation x² - 3x +2 = 0 or not.

Ans:  Here P(x) = x² - 3x +2=0, if x = -1 is a root of the equation P(-1) must be 0,

Put x = -1 in P(x)

P(1) = (-1)² - 3(-1) +2= 1 +3 +2=6$ \ne $0

Hence, x = -1 is not the solution of the given equation.

28. If x = $\frac{{\text{2}}}{{\text{3}}}$ is a solution of the quadratic equation 7x2+mx-3=0; Find the value of m.

Ans:  For P(x) = 7x² + mx - 3it is given x = $\frac{2}{3}$ is the solution of the given equation.

Put the given value of x in the given equation i.e P$\left( {\frac{2}{3}} \right)$ =0

P$\left( {\frac{2}{3}} \right)$ =  7$\left( {\frac{2}{3}} \right)$2 +m$\left( {\frac{2}{3}} \right)$- 3 = 0

⇒ 7$\left( {\frac{4}{9}} \right)$ +  m$\left( {\frac{2}{3}} \right)$ - 3 = 0

⇒ $\frac{1}{9}$ = - m$\left( {\frac{2}{3}} \right)$

⇒ m = -$\frac{1}{6}$ 

29. If x = -3 and x =$\frac{{\text{2}}}{{\text{3}}}$  are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.

Ans:  

For x = $\frac{2}{3}$ and x = -3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = $\frac{2}{3}$ and x = -3 in the given equation, we get

m$\left( {\frac{2}{3}} \right)$2+ 7$\left( {\frac{2}{3}} \right)$+ n = 0

⇒ m$\left( {\frac{4}{9}} \right)$+ $\left( {\frac{{14}}{3}} \right)$+ n = 0

⇒ 4m+ 9n+ 42 = 0                 ………. (i)

m(-3 )2+ 7($ - 3$)+ n = 0

⇒ 9m + n - 21 = 0

⇒n = 21 - 9m………..(2)

Put n in (1),

4m + 9(21-9m) + 42 = 0

⇒ m = 3

Substitute value of m in ( 2),

n = 21 - 9(3)

⇒ n = - 6

Thus, the value of m is 3 and the value of n is −6

30. If quadratic equation x² - (m + 1) x + 6 = 0 has one root as x = 3; find the value of m and the root of the equation.

Ans:  For p(x) = x² - (m + 1) x + 6 it is given x = 3 is the solution of the given equation. Put the given value of x in the given equation i.e p($3$) =0

p($3$) = (3)2 - (m + 1) 3+ 6 = 0

⇒ (9) - (3m + 3) + 6 = 0

⇒ 12 = 3m

⇒ m = 4

Substitute 4 for m in the equation,

x² - 5x + 6 = 0

⇒ x² - 3x -2x + 6 = 0

⇒ x(x-3) - 2(x-3) = 0

⇒ (x-3)(x-2) =0

Since, x-3=0 or x-2 = 0

Then x =3 or x =2

31. Given that 2 is a root of the equation 3x² - p(x + 1) = 0 and that the equation px² - qx + 9 = 0 has equal roots, find the values of p and q.

Ans:  For P(x) = 3x² - p(x + 1) = 0 it is given x = 2 is the solution of the given equation. Put the given value of x in the given equation i.e P(2) =0

P($3$) = 3(2)² - p(2 + 1) = 0

⇒ (27) - (3p) = 0

⇒ 27 = 3p

⇒ p = 9

a = 7, b = -9, c = 2

For Q(x) = px² - qx + 9 = 0 has equal roots the Discriminant of Q(x) must be 0.

a = p = 9, b = -q, c = 9

∴ Discriminant = b² - 4ac =0

D = (-q)² -4(9)(9) =0

$ \Rightarrow $(-q)² - 4(9)(9) =0

$ \Rightarrow $ (q)² -144 =0

$ \Rightarrow $ (q -12)(q+12) =0 

Hence, q -12 = 0 or q+12 = 0 

q = 12 or q = -12

∴ p =9 & q = $ \pm \;12$        

32. $\frac{x}{a}$ - $\frac{{a + b}}{x}$ = $\frac{{b\left( {a + b} \right)}}{{ax}}$       

Ans:  $\frac{x}{a}$ - $\frac{{a + b}}{x}$ = $\frac{{b\left( {a + b} \right)}}{{ax}}$

⇒ ax \[\left( {\frac{x}{a} - \frac{{a + b}}{x}} \right)\] = ax $\left( {\frac{{b\left( {a + b} \right)}}{{ax}}} \right)$

⇒ x2-a(a+b)=b(a+b)

⇒ x²=a(a+b)+b(a+b)

⇒ x²=(a+b)(a+b)

⇒ x²=(a+b)²

⇒ x = $ \pm $(a+b)

33. Solve the equation $\left( {\frac{{{\text{1200}}}}{{\text{x}}}{\text{ + 2}}} \right)$ (x -10) -1200 = 60

Ans:  $\left( {\frac{{1200}}{x} + 2} \right)$ (x -10) -1200 = 60

⇒ $\left( {\frac{{1200}}{x} + 2} \right)$ (x -10) =1260

⇒ $\left( {\frac{{600 + x}}{x}} \right)$ (x -10) =630

⇒ (600+x)(x -10) =630x

⇒ 600x - 6000 + x2-10x = 630x

⇒ x2 - 40x - 6000 = 0

⇒ x2 + 60x - 100x - 6000 = 0

⇒ x(x+60) - 100(x+60) = 0

⇒ (x-100)(x+60) =0

Hence, x -100=0      or x+60 =0 

             x=100          or x = -60

34. If -1 and 3 are the roots of x²+px+q=0 then find the values of p and q

Ans:  For x =-1 and x = 3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = -1 and x = 3 in the given equation, we get

(-1)²+p(-1)+q = 0

⇒1-p+q = 0 

⇒1+q = p ……. (1)

(3)2+p(3)+q = 0

⇒ 9+3p+q = 0 ………..(2)

Put p in (2)from (1),

9+3(1+q)+q = 0

⇒12+4q = 0

⇒ q = -3

Put value of q in (1)  

p = 1-3 

⇒ p = -2

Thus, the values of p and q are − 2 and 3 respectively.

Exercise 5(D)

1. Solve each of the following equations using the formula:

(i) x² - 6x =27 

Ans:  x² - 6x =27

Here, a =1, b=-6 , c=-27

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 6} \right)\; \pm \;\sqrt {{6^2} - 4\left( 1 \right)\left( { - 27} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{ - \left( { - 6} \right)\; \pm \;\sqrt {36 + 108} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{6\; \pm \;\sqrt {144} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{6 \pm 12}}{2}$= $\frac{{6\; + \;12}}{2}$or $\frac{{6\; - \;12}}{2}$= 9 or -3

(ii) x² - 10x +21=0

Ans:  x² - 10x +21 = 0

Here, a = 1, b = -10, c = 21

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 10} \right)\; \pm \;\sqrt {{{\left( { - 10} \right)}^2} - 4\left( 1 \right)\left( {21} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{10\; \pm \;\sqrt {100 - 84} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{10\; \pm \;\sqrt {16} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{10 \pm 4}}{2}$ = $\frac{{10\; + \;4}}{2}$ or $\frac{{10\; - \;4}}{2}$= 7 or 3    

(iii) x²+6x-10=0

Ans:  x² +6x -10=0

Here, a =1, b=6, c=-10

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( 6 \right)\; \pm \;\sqrt {{{\left( 6 \right)}^2} - 4\left( 1 \right)\left( { - 10} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{6\; \pm \;\sqrt {36 + 40} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{6\; \pm \;\sqrt {76} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{6\; \pm \;2\sqrt {19} }}{2}$= $3\; \pm \;\sqrt {19} $

(iv) x² +2x – 6 = 0

Ans:  x² + 6x -10 = 0

Here, a = 1, b = 2, c = -6

 Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

 ⇒ x = $\frac{{ - \left( 2 \right)\; \pm \;\sqrt {{{\left( 2 \right)}^2} - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{ - 2\; \pm \;\sqrt {4 + 24} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{ - 2\; \pm \;\sqrt {28} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{ - 2\; \pm \;2\sqrt 7 }}{2}$= $ - 1\; \pm \;\sqrt 7 $ 

(v) 3x²+ 2x - 1=0

Ans:  3x² +2x -1=0

Here, a =2, b=2, c=-1

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

  ⇒ x = $\frac{{ - \left( 2 \right)\; \pm \;\sqrt {{{\left( 2 \right)}^2} - 4\left( 2 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$

 ⇒ x  = $\frac{{ - 2\; \pm \;\sqrt {4 + 8} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{ - 2\; \pm \;\sqrt {12} }}{2}$

 ⇒ x = $\frac{{ - 2\; \pm 2\;\sqrt 3 }}{2}$= $ - 1\; \pm \;\sqrt 3 $ 

(vi) 2x²+7x+5=0

Ans:  2x²+7x+5=0

Here, a =2, b=7, c=5

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( 7 \right)\; \pm \;\sqrt {{7^2} - 4\left( 2 \right)\left( 5 \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{ - \left( 7 \right)\; \pm \;\sqrt {49 - 40} }}{4}$

 ⇒  x = $\frac{{ - 7 \pm \;\sqrt 9 }}{4}$

 ⇒  x = $\frac{{ - 7 \pm 3}}{4}$= $\frac{{ - 7\; + \;3}}{4}$or $\frac{{ - 7\; - \;3}}{4}$= 1 or $\frac{{ - 10}}{4}$

(vii) $\frac{2}{3}$ x = - $\frac{1}{6}$ x2 - $\frac{1}{3}$ 

Ans:  $\frac{2}{3}$x = -$\frac{1}{6}$x2 -$\frac{1}{3}$

 ⇒  4x = -x² -2

 ⇒ x²+4x+2=0

Here, a =1, b=4, c=2

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( 4 \right)\; \pm \;\sqrt {{4^2} - 4\left( 1 \right)\left( 2 \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{ - 4\; \pm \;\sqrt {16 - 8} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{ - 4\; \pm \;\sqrt 8 }}{2}$

⇒ x = $\frac{{ - 4\; \pm 2\;\sqrt 2 }}{2}$= $ - 2\; \pm \;\sqrt 2 $ 

(viii) $\frac{1}{{15}}$ x2 + $\frac{5}{3}$ = $\frac{2}{3}$ x 

Ans:  $\frac{1}{{15}}$ x² + $\frac{5}{3}$= $\frac{2}{3}$x

 ⇒ x² -10x+25=0

Here, a =1, b=-10, c=25

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

 ⇒ x = $\frac{{ - \left( { - 10} \right)\; \pm \;\sqrt {{{\left( { - 10} \right)}^2} - 4\left( 1 \right)\left( {25} \right)} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{10\; \pm \;\sqrt {100 - 100} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{10\; \pm \;\sqrt 0 }}{2}$

 ⇒ x = 5

(ix) x² - 6= 2$\sqrt 2 $x

Ans:  x² - 6= 2$\sqrt 2 $x

⇒x² - 2$\sqrt 2 $x -6 =0

Here, a =1, b=- 2$\sqrt 2 $ , c=-6

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - \;2\sqrt 2 } \right)\; \pm \;\sqrt {{{\left( { - \;2\sqrt 2 } \right)}^2} - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{\;2\sqrt 2 \; \pm \;\sqrt {8 + 24} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{\;2\sqrt 2 \; \pm \;\sqrt {32} }}{2}$

⇒ x = $\frac{{\;2\sqrt 2 \; \pm 4\sqrt 2 }}{2}$= $\sqrt 2 \; + 2\sqrt 2 $ or $\sqrt 2 \; - 2\sqrt 2 $ =3$\sqrt 2 $ or - $\sqrt 2 $

(x) $\frac{4}{x}$-3 = $\frac{5}{{2x + 3}}$

Ans:  For $\frac{4}{x}$-3= $\frac{5}{{2x + 3}}$

⇒$\frac{{4 - 3x}}{x}$= $\frac{5}{{2x + 3}}$

⇒ (4-3x)(2x+3)=5x

⇒ -6x2+8x-9x+12=5x

⇒ 6x2+6x-12=0

⇒ x2+x-2=0

Here, a =1, b=1, c=-2

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( 1 \right)\; \pm \;\sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 2} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{ - 1\; \pm \;\sqrt {1 + 8} }}{2}$

⇒ x = $\frac{{ - 1\; \pm \;\sqrt 9 }}{2}$

⇒ x = $\frac{{ - 1 \pm 3}}{2}$= $\frac{{ - 1\; + \;3}}{2}$or $\frac{{ - 1\; - \;3}}{2}$= 1 or -2

(xi) $\frac{{2x + 3}}{{x + 3}}$= $\frac{{x + 4}}{{x + 2}}$

Ans:  $\frac{{2x + 3}}{{x + 3}}$= $\frac{{x + 4}}{{x + 2}}$

⇒ (2x+3)(x+2)=(x+3)(x+4)

⇒ 2x²+4x+3x+6 = x²+4x+3x+12

⇒ x²-6 = 0

Here, a =1, b=0, c=-6

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( 0 \right)\; \pm \;\sqrt {{{\left( 0 \right)}^2} - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{\; \pm \;\sqrt {24} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{\; \pm \;\sqrt {24} }}{2}$

⇒ x = $\frac{{ \pm 2\;\sqrt 6 }}{2}$= $\; \pm \;\sqrt 6 $

(xii) $\sqrt 6 $x²-4x - 2$\sqrt 6 $=0

Ans: $\sqrt 6 $x²-4x - 2$\sqrt 6 $=0

Here, a=$\sqrt 6 $ , b=-4, c=-2$\sqrt 6 $

 Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 4} \right)\; \pm \;\sqrt {{{\left( { - 4} \right)}^2} - 4\left( { - 2\sqrt 6 } \right)\left( {\sqrt 6 } \right)} }}{{2\left( {\sqrt 6 } \right)}}$

⇒ x = $\frac{{4\; \pm \;\sqrt {16 + 48} }}{{2\left( {\sqrt 6 } \right)}}$

⇒ x = $\frac{{4\; \pm \;\sqrt {64} }}{{2\sqrt 6 }}$

⇒ x = $\frac{{4\; \pm 8}}{{2\sqrt 6 }}$ = $\frac{{4 + 8}}{{2\sqrt 6 }}$ or $\frac{{4 - 8}}{{2\sqrt 6 }}$ = $\frac{{12}}{{2\sqrt 6 }}$ or $\frac{{ - 4}}{{2\sqrt 6 }}$ = $\sqrt 6 $ or $\frac{{ - \sqrt 6 }}{3}$

(xiii) $\frac{{2x}}{{x - 4}}$+$\frac{{2x - 5}}{{x - 3}}$=8 $\frac{1}{3}$

Ans:  For $\frac{{2x}}{{x - 4}}$+$\frac{{2x - 5}}{{x - 3}}$=8 $\frac{1}{3}$

⇒ $\frac{{\left( {2x} \right)\left( {x - 3} \right) + \left( {2x - 5} \right)\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x - 3} \right)}}$= $\frac{{25}}{3}$

⇒3[2x(x-3)+(x-4)(2x-5)]=25(x-4)(x-3)

⇒ 3(4x2-19x+20)=25(x2-7x+12)

⇒ 13x2-118x+240=0

Here, a =13, b=-118, c=240

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

 ⇒ x = $\frac{{ - \left( { - 118} \right)\; \pm \;\sqrt {{{\left( { - 118} \right)}^2} - 4\left( {13} \right)\left( {240} \right)} }}{{2\left( {13} \right)}}$

 ⇒ x = $\frac{{118\; \pm \;\sqrt {1444} }}{{26}}$

 ⇒ x = $\frac{{118\; \pm \;38}}{{26}}$

 ⇒ x = $\frac{{118\; + \;36}}{{26}}$or $\frac{{118\; - \;36}}{{26}}$=6 or $\frac{{40}}{{13}}$

(xiv) $\frac{{x - 1}}{{x - 2}}$+$\frac{{x - 3}}{{x - 4}}$=3 $\frac{1}{3}$

Ans:  For $\frac{{x - 1}}{{x - 2}}$+$\frac{{x - 3}}{{x - 4}}$=3 $\frac{1}{3}$

⇒$\frac{{\left( {x - 1} \right)\left( {x - 4} \right) + \left( {x - 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}}$= $\frac{{10}}{3}$

⇒3[(x-1)(x-4)+(x-3)(x-2)]=10(x-2)(x-4)

⇒ 3(2x²-10x+10)=10(x²-6x+8)

⇒ 2x²-15x+25=0

a =2, b=-15, c=25

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 15} \right)\; \pm \;\sqrt {{{\left( { - 15} \right)}^2} - 4\left( 2 \right)\left( {25} \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{15\; \pm \;\sqrt {225 - 200} }}{4}$

⇒ x = $\frac{{15\; \pm \;\sqrt {25} }}{4}$

⇒ x = $\frac{{15\; \pm 5\;}}{4}$= $\frac{{15\; + 5}}{4}or\;\frac{{15\; - \;5}}{4} = \frac{{\;5}}{2}\;\;or\;5$

2. Solve each of the following equations for x and give, in each case, your answer correct to one decimal place:

(i) x²-8x+5=0

Ans:  x² - 8x+5=0

Here, a =1, b=-8, c=5

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 8} \right)\; \pm \;\sqrt {{{\left( { - 8} \right)}^2} - 4\left( 1 \right)\left( 5 \right)} }}{2}$

⇒ x = $\frac{{8\; \pm \;\sqrt {64 - 20} }}{2}$

⇒ x = $\frac{{8 \pm \;\sqrt {44} }}{2}$

⇒ x = $\frac{{8 \pm \;2\sqrt {11} }}{2}$= $4 \pm \;\sqrt {11} $ = $4 \pm 3.3$= 7.3 or 0.7

(ii) 5x² +10x - 3 =0

Ans:  5x² -10x-3=0

Here, a=5, b=-10, c=-3

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 10} \right)\; \pm \;\sqrt {{{\left( { - 10} \right)}^2} - 4\left( 5 \right)\left( { - 3} \right)} }}{{2\left( 5 \right)}}$

⇒ x = $\frac{{10\; \pm \;\sqrt {100 + 60} }}{{10}}$

⇒ x = $\frac{{10 \pm \;\sqrt {160} }}{{10}}$

⇒ x = $\frac{{10 \pm \;4\sqrt {10} }}{{10}}$= $1 \pm \frac{2}{5}\;\sqrt {10} $ = $4 \pm \frac{2}{5}\;\left( {3.2} \right)$= 0.3 or -2.3

3. Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:

(i) 2x² - 10x +5=0

Ans:  2x²-10x+5=0

Here, a=2, b=-10, c=5

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 10} \right)\; \pm \;\sqrt {{{\left( { - 10} \right)}^2} - 4\left( 2 \right)\left( 5 \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{10\; \pm \;\sqrt {100 - 40} }}{4}$

⇒ x = $\frac{{10 \pm \;\sqrt {60} }}{4}$

⇒ x = $\frac{{10 \pm \;2\sqrt {15} }}{4}$= $\frac{5}{2} \pm \frac{1}{2}\;\sqrt {15} $ = $2.5 \pm \frac{1}{2}\;\left( {3.9} \right)$= 4.44 or 0.56

(ii) 4x +$\frac{6}{x}$ +13=0

Ans:  4x +$\frac{6}{x}$ +6=0

⇒ 4x² +13x+6=0

Here, a=4 , b=13 , c=6

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( {13} \right)\; \pm \;\sqrt {{{\left( {13} \right)}^2} - 4\left( 4 \right)\left( 6 \right)} }}{{2\left( 4 \right)}}$

⇒ x = $\frac{{ - \left( {13} \right)\; \pm \;\sqrt {169 - 96} }}{8}$

⇒ x = $\frac{{ - 13 \pm \;\sqrt {73} }}{8}$

⇒ x = $\frac{{ - 13 + \;\sqrt {73} }}{8}$or $\frac{{ - 13 - \;\sqrt {73} }}{8}$ = $\frac{{ - 13 + 8.54}}{8}or\;\frac{{ - 13 - 8.54}}{8}\;$=-0.56 or -2.69

(iii) 4x² - 5x -3=0

Ans:  4x²-5x-3=0

Here, a =4, b=-5 , c=-3

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 5} \right)\; \pm \;\sqrt {{{\left( { - 5} \right)}^2} - 4\left( 4 \right)\left( { - 3} \right)} }}{{2\left( 4 \right)}}$

⇒ x = $\frac{{5\; \pm \;\sqrt {25 + 48} }}{8}$

⇒ x = $\frac{{5 \pm \;\sqrt {73} }}{8}$

⇒ x = $\frac{{5 + \;\sqrt {73} }}{8}$or $\frac{{5 - \;\sqrt {73} }}{8}$ = $\frac{{5 + 8.54}}{8}or\;\frac{{5 - 8.54}}{8}\;$=-0.44 or 1.69

(iv) x²-3x-9=0

Ans:  x²-3x-9=0

Here, a=1, b=-3, c=-9

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 3} \right)\; \pm \;\sqrt {{{\left( { - 3} \right)}^2} - 4\left( 1 \right)\left( { - 9} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{3\; \pm \;\sqrt {45} }}{2}$

⇒ x = $\frac{{3 \pm \;3\sqrt 5 }}{2}$

⇒ x = $\frac{{3 \pm \;3\left( {2.24} \right)}}{{10}}$= $\frac{{3 + 6.72}}{2}or\;\frac{{3 - 6.72}}{2}\;$=4.81 or -1.86

(v) x² - 5x - 10 = 0

Ans:  x²-5x-10=0

Here, a =1, b=-5, c=-10

 Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 5} \right)\; \pm \;\sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( { - 10} \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{5\; \pm \;\sqrt {25 + 40} }}{2}$

⇒ x = $\frac{{5 \pm \;\sqrt {65} }}{2}$

⇒ x = $\frac{{5 \pm \;8.06}}{2}$= $\frac{{5 + 8.06}}{2}or\;\frac{{5 - 8.06}}{2}\;$= 6.53 or -1.53

4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:

(i) 3x²-12x-1=0

Ans:  3x²-12x-1=0

Here, a =3, b=-12, c=-1

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 12} \right)\; \pm \;\sqrt {{{\left( { - 12} \right)}^2} - 4\left( 3 \right)\left( { - 1} \right)} }}{{2\left( 3 \right)}}$

⇒ x = $\frac{{12\; \pm \;\sqrt {144 + 12} }}{6}$

⇒ x = $\frac{{12 \pm \;2\sqrt {39} }}{6}$

⇒ x =  $2 \pm \frac{1}{3}\;\sqrt {39} $ = $2 \pm \frac{1}{3}\;\left( {6.2} \right)$= 4.082 or -0.082

(ii) x² - 16 x +6= 0

Ans:  x² -16x+6=0

Here, a =1, b=-16, c=6

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 16} \right)\; \pm \;\sqrt {{{\left( { - 16} \right)}^2} - 4\left( 1 \right)\left( 6 \right)} }}{{2\left( 1 \right)}}$

⇒ x = $\frac{{16\; \pm \;\sqrt {256 - 24} }}{2}$

⇒ x = $\frac{{16 \pm \;\sqrt {232} }}{2}$

⇒ x = $\frac{{16 \pm \;2\sqrt {58} }}{2}$= $8 \pm \sqrt {58} $ = $8 \pm \;\left( {7.615} \right)$= 15.615 or 0.384

(iii) 2x² + 11x + 4= 0

Ans:  2x² +11x+4=0

Here, a=2, b=11, c=4

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( {11} \right)\; \pm \;\sqrt {{{\left( {11} \right)}^2} - 4\left( 2 \right)\left( 4 \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{11\; \pm \;\sqrt {121 - 32} }}{4}$

⇒ x = $\frac{{11 \pm \;\sqrt {89} }}{4}$

⇒ x = $\frac{{11 \pm \;9.433}}{4}$= $\frac{{11 + \;9.433}}{4}$ or $\frac{{11 - \;9.433}}{4}$= -0.392 or -5.108

5. Solve:

(i) x⁴ - 2x² - 3 = 0

Ans:  x⁴ - 2x² - 3 = 0

Put y = x²

⇒ y² - 2y - 3 =0

⇒ y²-3y+y-3 =0

⇒ y(y-3)+1(y-3) =0

⇒ (y+1)(y-3) =0

Since, y-3 =0 or y+1 = 0

Then y = 3 or y =-1

⇒ x² = 3 or x² = -1(Rejected)                                 

⇒ x²-3 = 0

⇒ (x+ $\sqrt 3 $ )(x- $\sqrt 3 $ ) = 0

Since, x- $\sqrt 3 $ =0 or x + $\sqrt 3 $ = 0

Then x =$\sqrt 3 $ or x = -$\sqrt 3 $ 

(ii) x4 - 10x² +9 = 0

Ans:  x4 - 10x² +9 =0

Put    y = x²

⇒ y² - 10y +9 =0

⇒ y²-9y-y+9 =0

⇒ y(y-9)-1(y-9) =0

⇒ (y-1)(y-9) =0

Since, y-3 = 0 or y-1 = 0

Then y = 9 or y = 1

⇒ x² = 9 or x2=1                                

⇒ x²-9 = 0 or x²-1 = 0

⇒ (x+3)(x-3) = 0 or (x+1)(x-1) = 0

Since, x-3 = 0 or x+3 = 0 or x-1 = 0 or x+1 = 0

Then x = 3 or x = -3 or  x = 1  or x = -1

6. Solve:

(i) (x² - x)² + 5(x² - x) + 4 = 0

Ans:  (x² - x)² + 5(x² - x)+ 4=0

Put y = x²-x

⇒ y² +5y +4 =0

⇒ y² +4y+y+4 =0

⇒ y(y+4)+1(y+4) =0

⇒ (y+1)(y+4) =0

Since, y+4 =0 or y+1 = 0

Then y = -4 or y =-1

⇒ x²-x=-4 or  x²-x=-1                             

⇒ x²-x+4 = 0 or x2-x+1=0

For x²-x+4 = 0

a=1, b=-1, c=4

∴ Discriminant = b² - 4ac

                      D = (-1)² -4(1)(4)

                      D = 1-16

                      D =-15<0

Since D < 0. Therefore, the equation has no real roots.

For x²-x+1 = 0

a = 1, b = -1, c = 1

∴ Discriminant = b² - 4ac

                      D = (-1)² -4(1)(1)

                      D = 1-4

                      D = -3<0

Since D < 0. Therefore, the equation has no real roots.

(ii) (x² - 3x)² - 16(x² - 3x) - 36 =0

Ans: (x² -3 x)² -16(x² -3 x)-36=0

Put y = x²-3x

⇒ y² - 16y - 36 = 0

⇒ y² - 18y + 2y - 36 = 0

⇒ y(y-18)+2(y-18) = 0

⇒ (y+2)(y-18) =0

Since, y+4 = 0 or y-18 = 0

Then y = -2 or y =18

⇒ x²-3x+2 = 0 or x²-3x-18=0                             

For x²-3x+2 = 0 

⇒ x²-2x-x+2 =0

⇒ x(x-2)-1(x-2) =0

⇒ (x-2)(x-1) =0

Since, x-2 =0 or x-1 = 0

Then x =2 or x =1

For x²-3x-18 = 0

⇒ x²-6x+3x-18 =0

⇒ x(x-6)+3(x-6) =0

⇒ (x-6)(x+3) =0

Since, x-6=0 or x+3 = 0

Then   x =6    or x =-3

7.Solve  (i) $\sqrt {\frac{x}{{x - 3}}} $+$\sqrt {\frac{{x - 3}}{x}} $= $\frac{5}{2}$

Ans:  Put    y =$\sqrt {\frac{x}{{x - 3}}} $ 

⇒ y +$\frac{1}{y}$ = $\frac{5}{2}$

⇒ 2(y² +1) = 5y

⇒ 2y² – 5y + 2 = 0

⇒ 2y² – y – 4y + 2 = 0

⇒ y(2y-1)-2(2y-1) =0

⇒ (y-2)(2y-1) =0

Since, y-2=0 or 2y-1= 0

Then y =2 or y = $\frac{1}{2}$

$\;\sqrt {\frac{x}{{x - 3}}} $ = 2  or $\sqrt {\frac{x}{{x - 3}}} $=  $\frac{1}{2}$

⇒   $\frac{x}{{x - 3}}$= 4  or $\frac{x}{{x - 3}}$=  $\frac{1}{4}$                           

For $\frac{x}{{x - 3}}$= 4

⇒      x = 4(x-3) 

⇒      x=4x-12

⇒      3x =12

⇒      3x =12

⇒      x =4

(i) $\frac{x}{{x - 3}}$= $\frac{1}{4}$

⇒      4x = (x-3) 

⇒      4x=x-3

⇒      3x = -3

⇒      x = -1

(ii)$\frac{{2x - 3}}{{x - 1}}$-4$\frac{{x - 1}}{{2x - 3}}$=3

Ans:  Put y =$\frac{{2x - 3}}{{x - 1}}$ 

⇒    y -$\frac{4}{y}$ = 3

⇒    y² -4 = 3y

⇒   y² - 3y - 4 = 0

⇒   y² + y - 4y - 4 = 0

⇒   y(y+1)-4(y-4) =0

⇒   (y+1)(y-4) =0

Since, y-4 = 0 or y+1 = 0

Then y = 4 or y = -1

$\;\;\;\;\;\frac{{2x - 3}}{{x - 1}}\;$ = -1 or $\frac{{2x - 3}}{{x - 1}}$= 4

For $\frac{{2x - 3}}{{x - 1}}\;$ = -1

⇒      2x-3 = -1(x-1) 

⇒      2x-3=-x+1

⇒      3x =4

⇒      x = $\frac{4}{3}$

$\frac{{2x - 3}}{{x - 1}}$=  4

⇒      2x-3 = 4(x-1) 

⇒      2x-3=4x-4

⇒      2x = 1

⇒      x = $\frac{1}{2}$

(iii) $\frac{{3x + 1}}{{x + 1}}$+$\frac{{x + 1}}{{3x + 1}}$=$\frac{5}{2}$

Ans:  Put y =$\frac{{3x + 1}}{{x + 1}}$ 

⇒     y +$\frac{1}{y}$ = $\frac{5}{2}$

⇒    2(y²+1) = 5y

⇒    2y²-5y+2 = 0

⇒    2y²-y-4y+2 = 0

⇒    y(2y-1)-2(2y-1) = 0

⇒    (y-2)(2y-1) = 0

Since, y-2=0 or 2y-1= 0

Then y =2 or y = $\frac{1}{2}$

$\frac{{3x + 1}}{{x + 1}}\;$ = 2 or $\frac{{3x + 1}}{{x + 1}}$ = $\frac{1}{2}$

For  $\;\;\;\;\frac{{3x + 1}}{{x + 1}}\;$ = 2

⇒      3x+1 = 2(x+1) 

⇒      3x+1=2x+2

⇒      x =1

$\frac{{3x + 1}}{{x + 1}}\;$ = $\frac{1}{2}$

⇒      2(3x+1) = x+1 

⇒      6x+2=x+1

⇒      5x = -1

⇒      x =- $\frac{1}{5}$

8. Solve the equation 2x - $\frac{1}{x}$ =7. Write your answer correct to two decimal places.

Ans:  2x - $\frac{1}{x}$ =7

⇒       x + $\frac{1}{x}$= 7

⇒     2 x² - 1 = 7x 

⇒     2 x² -7x-1 = 0

Here, a = 2, b = -7, c = -1

Then     x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 7} \right)\; \pm \;\sqrt {{{\left( { - 7} \right)}^2} - 4\left( 2 \right)\left( { - 1} \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{7\; \pm \;\sqrt {49 + 8} }}{4}$

⇒ x = $\frac{{7 \pm \;\sqrt {57} }}{4}$

⇒ x = $\frac{{7 \pm \;7.55}}{4}$= $\frac{{7 + 7.55}}{4}or\;\frac{{7 - 7.55}}{4}\;$= 3.64 or -0.14

9. Solve the following equation and give your answer correct to 3 significant figures: 5x² - 3x - 4 = 0

Ans:  5x² - 3x - 4 = 0

Here, a=5, b=-3 , c=-4

 Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒               x = $\frac{{ - \left( { - 3} \right)\; \pm \;\sqrt {{{\left( { - 3} \right)}^2} - 4\left( 5 \right)\left( { - 4} \right)} }}{{2\left( 5 \right)}}$

 ⇒             x = $\frac{{3\; \pm \;\sqrt {9 + 80} }}{{10}}$

 ⇒             x = $\frac{{3 \pm \;\sqrt {89} }}{{10}}$

⇒             x = $\frac{{3 \pm \;9.434}}{{10}}$= $\frac{{3 + \;9.434}}{{10}}or\;\frac{{3 - \;9.434}}{{10}}\;$= 1.243 or -0.643

10. Solve for x using the quadratic formula. Write your answer correctly to two significant figures. (x - 1)² - 3x + 4 = 0

Ans:  (x - 1)² - 3x + 4 = 0

⇒ x² -2x+1- 3x + 4 =0

⇒ x² -5x+5 =0

Here, a =1, b=-5, c=5

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 5} \right)\; \pm \;\sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( 5 \right)} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{5\; \pm \;\sqrt {25 - 20} }}{2}$

 ⇒ x = $\frac{{5 \pm \;\sqrt 5 }}{2}$

⇒ x = $\frac{{5 \pm \;2.24}}{2}$ = $\frac{{5 + 2.24}}{2}\;or\;\frac{{5 - \;2.24}}{2}\;$ = 3.6 or 1.4

11. Solve the quadratic equation x² - 3(x + 3) = 0; Give your answer correct to two significant figures.

Ans:  x² - 3(x + 3) = 0

⇒ x²-3x-9 = 0

Here, a =1, b=-3, c=-9

 Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 3} \right)\; \pm \;\sqrt {{{\left( { - 3} \right)}^2} - 4\left( 1 \right)\left( { - 9} \right)} }}{{2\left( 1 \right)}}$

 ⇒ x = $\frac{{3\; \pm \;\sqrt {9 + 36} }}{2}$

⇒ x = $\frac{{3 \pm \;3\sqrt 5 }}{2}$

⇒ x = $\frac{{3 \pm \;6.708}}{2}$= $\frac{{3 + 6.708}}{2}\;or\;\frac{{3 - \;6.708}}{2}\;$= 4.9 or -1.9

Exercise 5(E)

Solve each of the Following equations 

1. $\frac{{{\text{2x}}}}{{{\text{x - 3}}}}$ + $\frac{{\text{1}}}{{{\text{2x + 3}}}}$ + $\frac{{{\text{3x + 9}}}}{{\left( {{\text{x - 3}}} \right)\left( {{\text{2x + 1}}} \right)}}$ = 0 Where x ≠ 3 x ≠ -$\frac{{\text{3}}}{{\text{2}}}$

Ans:  For $\frac{{2x}}{{\;x - 3}}$+$\frac{1}{{2x + 3}}$+ $\frac{{3x + 9}}{{\left( {x - 3} \right)\left( {2x + 1} \right)}}$=0

⇒$\frac{{2x\left( {2x + 3} \right) + \left( {x - 3} \right) + 3x + 9}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}$

⇒$2x\left( {2x + 3} \right) + \left( {x - 3} \right) + 3x + 9$=0

⇒ 4x²+6x+x-3+3x+9=0

⇒ 4x²+10x+6=0

⇒ 2x²+5x+3=0

⇒ 2x²+2x+3x+3=0

⇒ 2x(x+1)+3(x+1)=0

⇒ (x+1)(2x+3)=0

Since, x+1=0 or 2x+3 = 0

⇒ x=-1 or x=-$\frac{3}{2}$(rejected)

2. (2x+3)²=81

Ans:  (2x+3)²=81

⇒    (2x+3)²-81 =0 

⇒     (2x+3)²-(9)2 =0

⇒      (2x+12)(2x-6) =0

Since, 2x+12 =0 or 2x-6 = 0

⇒ 2x = -12 

⇒ 2x=6

Then x =-6 or x =3

3. a²x²-b² = 0

Ans:  a²x²-b²=0

⇒ (ax)²-b²=0

⇒ (ax+b)(ax-b) =0

Since, ax+b =0 or ax-b= 0

⇒ x=-$\frac{b}{a}$  x=$\frac{b}{a}$

Then x =-$\frac{b}{a}$    or x =$\frac{b}{a}$

4. x² - $\frac{{11}}{4}$ x + $\frac{{15}}{8}$ = 0

Ans:  x² - $\frac{{11}}{4}$ x + $\frac{{15}}{8}$ = 0

⇒      8x² - 22 x + 15 = 0

⇒     8x² -12x-10x + 15 = 0

⇒     4x(2x-3) - 5(2x-3) = 0

⇒     (4x-5)(2x-3) =0

Since, 4x-5 =0 or 2x-3 = 0

Then x =$\frac{5}{4}$ or x = $\frac{3}{2}$

5. x + $\frac{4}{x}$ = - 4

Ans:  x + $\frac{4}{x}$ = - 4

⇒ x² + 4 = -4x

⇒ x² + 4x + 4 = 0

⇒ (x+2)² = 0

Since, x+2 = 0

Then x = -2 

6. 2x⁴ - 5x² +3 = 0

Ans:  2x⁴ - 5x² +3 = 0

Put y = x²

⇒ 2y² - 5y +3 = 0

⇒ 2y² - 2y - 3y + 3 = 0

⇒ 2y(y - 1) - 3(y - 1) = 0

⇒ (y - 1)(2y - 3) = 0

Since, 2y - 3 = 0 or y - 1 = 0

Then y = $\frac{3}{2}$ or y = 1

⇒ x² = $\frac{3}{2}$ or x2 = 1                                 

⇒ x² - $\frac{3}{2}$ = 0 or x2 - 1 = 0

⇒ \[\left( {x + \sqrt {\frac{3}{2}} } \right)\]\[\left( {x + \sqrt {\frac{3}{2}} } \right)\] = 0 or (x+1) (x-1) = 0

Since, x - $\sqrt {\frac{3}{2}} $ = 0 or x +$\sqrt {\frac{3}{2}} $ = 0 or x - 1 = 0 or x + 1 = 0 

Then x = $\sqrt {\frac{3}{2}} $ or x = -$\sqrt {\frac{3}{2}} $ or x = 1 or x = -1

7. x⁴ - 2x² - 3 = 0

Ans:  x⁴ - 2x² - 3 = 0

Put y = x²

⇒     y² - 2y - 3 =0

⇒     y²-3y+y-3 =0

⇒     y(y-3)+1(y-3) =0

⇒     (y+1)(y-3) =0

Since, y-3 =0 or y+1 = 0

Then y = 3 or y =-1

⇒     x²=3    or  x2 = -1      (Rejected)                                 

⇒     x²-3 = 0

⇒     (x+$\sqrt 3 $)(x-$\sqrt 3 $) =0

Since, x-$\sqrt 3 $ =0 or x+$\sqrt 3 $ =0

Then x=$\sqrt 3 $ or x= -$\sqrt 3 $

8. Solve 9 $\left( {{x^2} + \frac{{1\;}}{{\;{x^2}}}} \right)$ - 9 $\left( {x + \frac{1}{x}} \right)$- 52 = 0

Ans:  9 $\left( {{x^2} + \frac{{1\;}}{{\;{x^2}}}} \right)$ - 9 $\left( {x + \frac{1}{x}} \right)$- 52 = 0

Let x + $\frac{1}{x}$ = y

Squaring on both sides,

x² + $\frac{{1\;}}{{\;{x^2}}}$ + 2 = y²

⇒ x² + $\frac{{1\;}}{{\;{x^2}}}$= y²-2 

Hence the equation becomes,

⇒     9(y²-2)-9y-52 =0

⇒     9y²-9y-70 =0

⇒     9y²-30y+21y-70 =0

⇒     3y(3y-10)+7(3y-10) =0

⇒     (3y+7)(3y-10) =0

Since, 3y+7 =0 or 3y-10 = 0

⇒ y = - $\frac{7}{3}$ or y = $\frac{{10}}{3}$

x + $\frac{1}{x}$ = - $\frac{7}{3}$ or x + $\frac{1}{x}$ = $\frac{{10}}{3}$

For $\;\;\;x + \frac{1}{x} =  - \;\frac{7}{3}\;$ 

⇒     3( x2+1)=-7x

⇒      3x2+7x+3=0

Then         x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

  ⇒             x = $\frac{{ - \left( 7 \right)\; \pm \;\sqrt {{{\left( 7 \right)}^2} - 4\left( 3 \right)\left( 3 \right)} }}{{2\left( 3 \right)}}$

 ⇒             x  = $\frac{{ - 7\; \pm \;\sqrt {49 - 36} }}{6}$           

 ⇒              x  = $\frac{{ - 7\; \pm \;\sqrt {13} }}{6}$

x+$\frac{1}{x}$ =$\frac{{10}}{3}$

⇒ 3x2-10x+3 =0

⇒3x2-9x-x+3 =0

⇒3x(x-3)-1(x-3) =0

⇒ (x-3)(3x-1) =0

Hence, x -3=0          or 3x-1 =0 

             x=3          or x = $\frac{1}{3}$

9.Solve 2(x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$) - (x+$\frac{1}{x}$)=11

Ans:  2(x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$) -(x+$\frac{1}{x}$)-11=0

Let  x+$\frac{1}{x}$=y

Squaring on both sides,

x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$+2 = y2

⇒x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$= y2-2 

Hence the equation becomes,

⇒      2(y2-2)-y-11 =0

⇒      2y2-y-15 =0

⇒      2y2-6y+5y-15 =0

⇒      2y(y-3)+5(y-3) =0

⇒      (2y+5)(y-3) =0

Since,  2y+5 =0 or y-3 = 0

⇒       y =- $\frac{5}{2}$or  y =3

 x+$\frac{1}{x}$=- $\frac{5}{2}$or  x+$\frac{1}{x}$ =3

For  $\;\;\;x + \frac{1}{x} =  - \;\frac{5}{2}\;$ 

⇒     2( x2+1)=-5x

⇒      2x2+5x+2=0

⇒      2x2+4x+x+2=0

⇒      2x(x+2)+1(x+2) =0

⇒      (2x+1)(x+2) =0

Hence, x +2=0          or  2x+1 =0 

             x=-2          or x =- $\frac{1}{2}$

$\;\;\;x + \frac{1}{x} = 3\;$ 

⇒      x2+1=3x

⇒      x2-3x+1=0

Then         x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

  ⇒             x = $\frac{{ - \left( { - 3} \right)\; \pm \;\sqrt {{{\left( { - 3} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$

 ⇒             x  = $\frac{{3\; \pm \;\sqrt {9 - 4} }}{2}$           

 ⇒              x  = $\frac{{3\; \pm \;\sqrt 5 }}{2}$

10.Solve (x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$) -3 (x+$\frac{1}{x}$)-2=0

Ans:  (x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$) -3(x+$\frac{1}{x}$)-2=0

Let  x+$\frac{1}{x}$=y

Squaring on both sides,

x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$+2 = y2

⇒x2 +$\frac{{1\;}}{{\;{x^2}\;\;}}$= y2-2 

Hence the equation becomes,

⇒      (y2-2)-3y-2 =0

⇒      y2-3y-4 =0

⇒      y2-4y+y-4 =0

⇒      y(y-4)+1(y-4) =0

⇒      (y+1)(y-4) =0

Since,  y+1 =0 or y-4 = 0

⇒         y =-1   or  y =4

 x+$\frac{1}{x}$=-1or  x+$\frac{1}{x}$ =4

For  $\;\;\;x + \frac{1}{x} =  - 1\;$ 

⇒     ( x2+1)=-x

⇒      x2+x+1=0

∴ Discriminant = b2 - 4ac

                         = (1)2 -4(1)(1)

                         = 1-4

                         = -3 < 0 

Since D <0. Therefore, the equation has no real roots.

 x+$\frac{1}{x}$ =4

⇒ x2-4x+1 =0

Then         x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

  ⇒             x = $\frac{{ - \left( { - 4} \right)\; \pm \;\sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$

 ⇒             x  = $\frac{{4\; \pm \;\sqrt {16 - 4} }}{2}$           

 ⇒              x  = $\frac{{4\; \pm \;\sqrt {12} }}{2}$= $\frac{{4\; \pm 2\;\sqrt 3 }}{2}$= 2$ \pm \sqrt 3 $

11.Solve (x2 +5x+4)(x2+ 5x+6)=120

Ans:  (x2 +5x+4)(x2+ 5x+6)=0

Put    y = x2+5x+4

        (y)(y+2)=120

⇒     y2 +2y-120 =0

⇒    y2 +12y-10y-120 =0

⇒      y(y+12)-10(y+12) =0

⇒      (y-10)(y+12) =0

Since,  y-10=0 or y+12 = 0

Then   y = 10    or y=-12

⇒      x2 +5x+4 =10    or  x2 +5x+4=-12

 ⇒      x2 +5x-6 =0    or  x2 +5x+16=0                          

For  x2 +5x-6 =0

⇒      x2+6x-x-6 =0

⇒      x(x+6)-1(x+6) =0

⇒      (x-1)(x+6) =0

Since,  x+6 =0 or x+1 = 0

Then   x =-6    or x =1

For  x2 +5x+16=0       

a =1 , b=5, c=16

∴ Discriminant = b2 - 4ac

                      D = (5)2 -4(1)(16)

                      D = 25-64

                      D =-39<0

Since D < 0. Therefore, the equation has no real roots.

12. Solve each of the following equations, giving answer upto two decimal places.

(i) x² - 5x -10=0

Ans:  2x² -10x+5=0

Here, a = 2, b = -10, c = 5

Then     x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( { - 10} \right)\; \pm \;\sqrt {{{\left( { - 10} \right)}^2} - 4\left( 2 \right)\left( 5 \right)} }}{{2\left( 2 \right)}}$

⇒ x = $\frac{{10\; \pm \;\sqrt {100 - 40} }}{4}$

⇒ x = $\frac{{10 \pm \;\sqrt {60} }}{4}$

⇒ x = $\frac{{10 \pm \;2\sqrt {15} }}{4}$= $\frac{5}{2} \pm \frac{1}{2}\;\sqrt {15} $ = $2.5 \pm \frac{1}{2}\;\left( {3.9} \right)$= 4.44 or 0.56

(ii) 3x² - x - 7 =0

Ans:  4x + $\frac{6}{x}$ + 6 = 0

 ⇒ 4x² + 13x + 6 = 0

Here, a = 4, b = 13, c = 6

Then x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ x = $\frac{{ - \left( {13} \right)\; \pm \;\sqrt {{{\left( {13} \right)}^2} - 4\left( 4 \right)\left( 6 \right)} }}{{2\left( 4 \right)}}$

⇒ x = $\frac{{ - \left( {13} \right)\; \pm \;\sqrt {169 - 96} }}{8}$

⇒ x = $\frac{{ - 13 \pm \;\sqrt {73} }}{8}$

⇒ x = $\frac{{ - 13 + \;\sqrt {73} }}{8}$or $\frac{{ - 13 - \;\sqrt {73} }}{8}$ = $\frac{{ - 13 + 8.54}}{8}or\;\frac{{ - 13 - 8.54}}{8}\;$=-0.56 or -2.69

13. Solve. $\left( {\frac{{\text{x}}}{{{\text{x + 2}}}}} \right)$2 - 7$\left( {\frac{{\text{x}}}{{{\text{x + 2}}}}} \right)$ + 12 = 0, x$ \ne $2

Ans:  ${\left( {\frac{x}{{x + 2}}} \right)^2}$- 7 $\left( {\frac{x}{{x + 2}}} \right)$ + 12 = 0, x$ \ne $2

Let y = $\frac{x}{{x + 2}}$

⇒      y²-7y+12 =0

⇒      y²-4y-3y+12 =0

⇒      y(y-4)-3(y-4) =0

⇒      (y-4)(y-3) =0

Since, y-4=0 or y-3 = 0

Then, y =4 or y =3

$\frac{x}{{x + 2}}$= 4 or $\frac{x}{{x + 2}}$= 3

⇒ $\frac{x}{{x + 2}}$= 4 or $\frac{x}{{x + 2}}$ = 3                           

For $\frac{x}{{x + 2}}$ = 4

⇒      x = 4(x+2) 

⇒      x = 4x+8

⇒      -3x = 8

⇒      x = - $\frac{8}{3}$

For $\frac{x}{{x + 2}}$ = 3

⇒      x = 2(x+2) 

⇒      x=3x+6

⇒      -2x =6

⇒      x = -3

14. Solve:

(i) x² - 11x - 12 =0; when x $ \in $N

Ans:  x²-11x-12 =0

⇒      x²-12x+x-12 =0

⇒      x(x-12)+1(x-12) =0

⇒      (x+1)(x-12) =0

Since, x-12=0 or x+1 = 0

Then, x =12   or x =-1

As x $ \in $N x =12

(ii) x² - 4x - 12 =0; when x$ \in $ I

Ans:  x² - 4x - 12 = 0

⇒      x² - 2x + 6x - 12 = 0

⇒      x(x-2)+6(x-2) = 0

⇒      (x+6)(x-2) = 0

Since,  x-2 = 0 or x+6 = 0

Then, x =2 or x =-6

As x $ \in $I 

x =2 or x =-6

(iii) 2x² - 9x + 10 = 0; when x $ \in $Q

Ans:  2x²-9x+10 =0

⇒      2x²-4x-5x-10 =0

⇒      2x(x-2)-5(x-2) =0

⇒      (2x-5)(x-2) =0

Since, 2x-5=0 or x-2 = 0

Then, x =$\frac{5}{2}$  or x =2

As x $ \in $Q 

x =$\frac{5}{2}$ or x = 2

15. (a+b)x²-(a+b)x-6 = 0

Ans:  (a+b)2x²-(a+b)x-6 = 0

⇒     (a+b)2x²-3(a+b)x+2(a+b)x-6 = 0

⇒      (a+b)x[(a+b)x-3]+2[(a+b)x-3] = 0

⇒      [(a+b)x-3] [(a+b)x+2] = 0

Since, (a+b)x-3 = 0 or (a+b)x+2 = 0

Then x =$\frac{3}{{a + b}}$ or x = - $\frac{2}{{a + b}}$ 

16. $\frac{1}{p}$ + $\frac{1}{q}$ + $\frac{1}{x}$ = $\frac{1}{{x + p + q}}$

Ans:  For $\frac{1}{p}$+$\frac{1}{q}$+$\frac{1}{x}$= $\frac{1}{{x + p + q}}$

⇒$\frac{1}{p}$+$\frac{1}{q}$+$\frac{1}{x}$- $\frac{1}{{x + p + q}}$ = 0

⇒$\frac{{q + p}}{{pq}}$+ $\frac{{x + p + q - x}}{{\left( x \right)\left( {x + p + q} \right)}}$= 0

⇒ (q+p)(x)(x+p+q)+(p+q)(pq)=0

⇒ (p+q)[x²+(p+q)x+pq]

⇒ x²+px+qx+pq = 0

⇒      x(x+p)+q(x+p) = 0

⇒      (x+p)(x+q) =0

Since, x+p = 0 or x+q = 0

Then   x=-p or x=-q 

17. Solve: (i) (x)(x + 1)+ (x+2)(x+3)=42

Ans:  (x)(x +1)+(x+2)(x+3) = 42

⇒      x² + x + x2+2x+3x+6 = 42

⇒      2x²+6x-36 = 0

⇒      x²+3x-18 = 0

⇒      x²+6x-3x-18 = 0

⇒      x(x+6)-3(x+6) = 0

⇒      (x+6)(x-3) = 0

Since, x-6 = 0 or x-3 = 0

Then x=-6 or x=3 

(ii)$\frac{1}{{x + 1}}$-$\frac{2}{{x + 2}}$= $\frac{3}{{x + 3}}$- $\frac{4}{{x + 4}}$

Ans:  For $\frac{1}{{x + 1}}$-$\frac{2}{{x + 2}}$= $\frac{3}{{x + 3}}$- $\frac{4}{{x + 4}}$

⇒$\frac{{x + 2 - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}$= $\frac{{3\left( {x + 4} \right) - 4\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}$

⇒$\frac{{ - x}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}$= $\frac{{ - x}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}$

⇒ -x(x+3)(x+4)=-x(x+1)(x+2)

⇒ -x(x2+7x+12)=-x(x2+3x+2)

⇒ -x(4x+10)=0

Since,  x=0 or 4x+10 = 0

⇒ x=0 or x=-$\frac{5}{2}$

18. For each equation, given below, find the value of m so that the equation has equal roots. Also, the solution of each equation

(i) (m-3)x² -4x +1 =0

Ans:  Since, the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b2 - 4ac =0

   Here, a=m-3, b=-4, c = 1

    D = (-4)² -4(m-3)(1) = 0

⇒ 16-4m+12 =0

⇒ 16-4m+12 =0

⇒ m= 7

4x²-4x+1 = 0

⇒ (2x-1)² =0

Since, 2x-1 =0 

Then x =$\frac{1}{2}$

(ii) 3x²+12x+(m+7)=0

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac = 0

   Here, a = 3, b=12, c=m+7

   D = (12)² -4(3)(m+7) = 0

⇒ 144 -12m-84= 0

⇒ 60 = 12m

⇒ m = 5

3x²+12x+12 =0

⇒ x²+4x+4 =0

⇒      (x+2)² = 0

Since, x+2 = 0 

Then x =-2

(iii) x²-(m+2)x +(m+5)=0

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b2 - 4ac =0

   Here, a = 1, b = -(m+2), c = m+5

    D = (m+2)² -4(m+5)(1) = 0

⇒ (m)² + 4m + 4 - 4(m+5) = 0

⇒ m² -16= 0

⇒ (m-4)(m+4)= 0

Since, m-4 =0 or m+4 = 0

Then m =4 or m =-4

Thus, the values of m are 4 and -4

19. Without solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots. px² - 4x +3 = 0

Ans:  Since, the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac = 0

Here, a = p, b = -4, c = 3

D = (-4)2 -4(p)(3) = 0

⇒ 16-12p = 0

⇒ p= $\frac{4}{3}$

20.Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.x² -2 (m-1)x +(m+5) =0

Ans:  Since, the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac =0

   Here, a = 1, b = 2(m -1), c = m+5

    D = (2m-2)² -4(m+5)(1) = 0

⇒ (2m)² -8m+4-4(m+5)= 0

⇒ 4m² - 12m – 16 = 0

⇒ m² - 3m – 4 = 0

⇒ m² - 4m + m – 4 = 0

⇒ m(m-4) + 1(m-4) = 0

⇒ (m-4)(m+1)= 0

Since, m-4 = 0 or m+1 = 0

Then m = 4 or m = -1

Thus, the values of m are 4 and -1 

Exercise 5(F)

1. (i)Solve : (x+5)(x-5)=24 

Ans:   (x+5)(x-5)=24

⇒      x²-24 =25

⇒      x²-49=0

⇒      (x+7)(x-7) =0

Since, x-7 =0 or x+7 = 0

Then x =7   or x =-7

(ii) 3x² -2( $\sqrt 6 $ )x+2=0

Ans:  3x² -2($\sqrt 6 $ )x+2=0

⇒     3x² - ( $\sqrt 6 $ )x-( $\sqrt 6 $ )x+2 = 0

⇒      $\sqrt 3 $x($\sqrt 3 $x -$\sqrt 2 $ )- $\sqrt 2 $( $\sqrt 3 $ x- $\sqrt 2 $) = 0

⇒       ( $\sqrt 3 $ x - $\sqrt 2 $ )² = 0

Since, $\sqrt 3 $ x - $\sqrt 2 $ = 0

Then x =$\sqrt {\frac{2}{3}} $

(iii) 3 $\sqrt 2 $ x² -5x+ $\sqrt 2 $ = 0

Ans:  3 $\sqrt 2 $ x²-5x+ $\sqrt 2 $ = 0

⇒     3 $\sqrt 2 $ x² - 6x+x+ $\sqrt 2 $ =0

⇒       $3\sqrt 2 $x(x -$\sqrt 2 $)+1(x-$\sqrt 2 $)=0

⇒        (3$\sqrt 2 $x -1) (x -$\sqrt 2 $) = 0

Since, 3$\sqrt 2 $x -1 =0 or x -$\sqrt 2 $ =0

Then   x = $\frac{1}{{3\sqrt 2 }}$ or x =$\sqrt 2 $

(iv)2x - 3 = $\sqrt {2\;{x^2}\; - \;2x\; + \;21} $

Ans:  Squaring on both the sides, we get

(2x - 3)² = 2x2 - 2x + 21

⟹ 4x² - 12x + 9 = 2x2 - 2x + 21

⟹ 2x² - 10x - 12 = 0

⟹ x² - 5x - 6 = 0 …..Dividing equation by 2

⟹ x² - 6x + x - 6 = 0

⟹ x(x - 6) + 1(x - 6) = 0

⟹ (x - 6)(x + 1) = 0

⟹ (x - 6) = 0 or (x + 1) = 0

⟹ x = 6 or x = -1

2. One root of the quadratic equation 8x2+ mx+ 15 = 0 is$\frac{3}{4}$ . Find the value of m. Also, find the other root of the equation. 

Ans:  For quadratic equation is 8x2 + mx + 6 = 0 One of the roots of is $\frac{3}{4}$, so it satisfies it

8 ${\left( {\frac{3}{4}} \right)^2}$+ m $\left( {\frac{3}{4}} \right)$ + 15 = 0

⇒ 8 $\left( {\frac{9}{{16}}} \right)$ + m $\left( {\frac{3}{4}} \right)$ + 15 = 0

⇒ $\frac{9}{2}$ + m $\left( {\frac{3}{4}} \right)$ + 15 = 0

⇒ m $\left( {\frac{3}{4}} \right)$ = - $\frac{{39}}{2}$

⇒ m= -26

So the equation becomes 8x²-26x+15 = 0

8x²-26x+15 = 0

⇒    8x²-20x-6x+15 = 0

⇒    4x(2x-5)-3(2x-5) = 0

⇒    (4x-3)(2x-5) = 0

Since, (4x-3)=0 or 2x-5 = 0

Then   x =$\frac{3}{4}$     or x =$\frac{5}{2}$

Hence the other root is $\frac{5}{2}$

3. One root of the quadratic equation x²+ (3-2a)x-6a = 0 is -3, find its other root. 

Ans:  For quadratic equation is x²+ (3-2a)x - 6a = 0 One of the roots of  is -3, so it satisfies it

⇒    x² + 3x - 2ax - 6a = 0

⇒    x(x+3)-2a(x+3) = 0

⇒    (x+3)(x-2a) = 0

Since, (x+3)=0 or x-2a = 0

Then x = -3 or x = 2a

Hence, the other root is 2a.

4. If p-15 =0 and 2x²+px+25 = 0 find the values of x. 

Ans:  Given p-15 =0 i.e p =15

So, the given quadratic equation becomes,

2x² + 15x + 25 = 0

⇒      2x²+10x+5x+25 =0

⇒      2x(x+5)+5(x+5) =0

⇒      (2x+5)(x+5) =0

Since, 2x+5=0 or x+5 = 0

Then x = -$\frac{5}{2}$ or  x = -5.

5. Find the solution of the equation 2x2-mx-25n = 0; if and m+5 =0 & n-1=0 

Ans:  Given quadratic equation is 2x²-mx-25n = 0

Also, given and m+5 =0 & n-1=0 

⇒   m = -5 and n =1

So, the given quadratic equation becomes,

2x²+15x-25 = 0

⇒      2x2+10x-5x-25 =0

⇒      2x(x+5)-5(x+5) =0

⇒      (2x-5)(x+5) =0

Since, 2x-5=0 or x+5 = 0

Then x = $\frac{5}{2}$ or x = -5.

Hence, the solution of given quadratic equation are $\frac{5}{2}$ and -5.

6. If m and n are roots of the equation $\frac{1}{x}$-$\frac{1}{{x - 2}}$= 3 where x ≠ 0 and x ≠ 2; find m × n.

Ans:  Given quadratic equation is $\frac{1}{x}$-$\frac{1}{{x - 2}}$= 3

For $\frac{4}{{x + 2}}$-$\frac{1}{{x + 3}}$= 3

⇒$\frac{{x - 2 - x}}{{\left( x \right)\left( {x - 2} \right)}}$=3 

⇒$\frac{{ - 2}}{{\left( x \right)\left( {x - 2} \right)}}$=3

⇒-2=3(x2-2x)

⇒ 3x2-6x+2=0

Here the roots of the equation aare m and n,

m x n = $\frac{{constant\;term\;}}{{Coefficient\;of\;{x^2}\;\;}}$= $\frac{2}{3}$

7. Solve, using formula: x² + x - (a+2)(a+1) = 0

Ans:  Given quadratic equation is x² + x - (a+2)(a+1)

Using quadratic formula,

Here, A = 1, B = 1, C = -(a+2)(a+1)

Then x = $\frac{{ - B\; \pm \;\sqrt {{B^2} - 4AC} }}{{2A}}$

⇒            x = $\frac{{ - \left( 1 \right)\; \pm \;\sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right) - \left( {a + 2} \right)\left( {a + 1} \right)} }}{{2\left( 1 \right)}}$

 ⇒             x = $\frac{{ - 1\; \pm \;\sqrt {1 + 4({a^2} + 3a + 2)} }}{{2\left( 1 \right)}}$

 ⇒             x = $\frac{{ - 1\; \pm \;\sqrt {4{a^2} + 12a + 9} }}{{2\left( 1 \right)}}$

⇒             x = $\frac{{ - 1\; \pm \;\sqrt {{{(2a + 3)}^2}} }}{2}$

⇒             x = $\frac{{ - 1\; \pm \left( {2a + 3} \right)}}{2}$

 ⇒             x = $\frac{{ - 1\; + \;2a + 3}}{2}$or $\frac{{ - 1\; - \;2a - 3}}{2}$

⇒              x = a + 1 or x = -a - 2 = -(a + 2)

8. Solve the quadratic equation 8x²-14x+3 =0

(i) When x ∈ I (integers)

(ii) When x ∈ Q (rational numbers) 

Ans:  8x²-14x+3 =0

⇒      8x²-12x-2x+3 =0

⇒      4x(2x-3)-1(2x-3) =0

⇒      (4x-1)(2x-3) =0

Since, 4x-1=0 or 2x-3 = 0

Then x = $\frac{1}{4}$ or x =$\frac{3}{2}$

(i) When x ∈ I, the equation has no roots

(ii) When the roots of are equation x = $\frac{1}{4}$& x =$\frac{3}{2}$

9. Find the value of m for which the equation(m+4)x2 +(m+1)x +1 =0 has real and equal roots. 

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b2 - 4ac =0

   Here a =m+1, b=(m+1), c = 1

    D = (m+1)² -4(m+4)(1) = 0

⇒ (m)² +2m+1-4(m+4)= 0

⇒ m² – 2m – 15 = 0

⇒ m² -5m+3m-15 = 0

⇒ m(m-5)+3(m-5) = 0

⇒ (m-5)(m+3) = 0

Since, m-5 = 0 or m+3 = 0

Then m = 5 or m =-3

Thus, the values of m are 5 and -3

10. Find the values of m for which equation 3x² + mx + 2 = 0 has equal roots. Also, find the roots of the given equation. 

Ans:  Since the equation has equal roots. Therefore, the discriminant will be zero.

∴ Discriminant = b² - 4ac = 0

   Here, a = 3, b = m, c = 2

    D = (m)² - 4(3)(2) = 0

⇒ m² – 24 = 0

⇒ (m)² - (2 $\sqrt 6 $ )2 = 0

⇒ (m - 2 $\sqrt 6 $ )(m + 2 $\sqrt 6 $ ) = 0

Since, m-2$\sqrt 6 $ =0 or m+2$\sqrt 6 $= 0

Then m =2$\sqrt 6 $ or m =-2$\sqrt 6 $

Thus, the values of m are 2$\sqrt 6 $ and -2$\sqrt 6 $

For, m = 2$\sqrt 6 $,

3x² +2$\sqrt 6 $x+2 =0

⇒      ($\sqrt 2 $x)² + 2$\sqrt 3 \sqrt 2 $ x + ( $\sqrt 2 $ )²

⇒      ( $\sqrt 3 $ x + $\sqrt 2 $ )² = 0

⇒      Since, $\sqrt 3 $ x + $\sqrt 2 $ = 0 Then x = - $\frac{{\sqrt 2 }}{{\sqrt 3 }}$

For, m = -2$\sqrt 6 $,

  3x² -2$\sqrt 6 $x+2 =0

⇒      ( $\sqrt 2 $ x)² -2$\sqrt 3 \sqrt 2 $ x + ( $\sqrt 2 $ )²

⇒      ( $\sqrt 3 $ x - $\sqrt 2 $ )² = 0

⇒      Since, $\sqrt 3 $ x - $\sqrt 2 $ = 0 Then x = $\frac{{\sqrt 2 }}{{\sqrt 3 }}$

11. Find the value of k for which equation 4x² +8x+-k=0 has real roots. 

Ans:  For Given quadratic equation to have real roots its discriminant is greater than or equal to zero,

∴ Discriminant = b² - 4ac =0

   Here  a = 3, b = m, c = 2

    D = (8)² -4(4)(-k) $ \geqslant $ 0

⇒ 64 + 16k $ \geqslant $ 0

⇒ 16k $ \geqslant $ -64

⇒ k $ \geqslant $ -4

Hence, the given quadratic equation has real roots for k $ \geqslant $-4.

12. Find, using quadratic formula, the roots of the following quadratic equations, if they exist

(i) 3x² - 5x + 2 = 0

Ans:  3x² - 5x +2 = 0

Here, a = 3, b = -5 , c = 2

 Then         x = $\frac{{ - b\; \pm \;\sqrt {{b^2} - 4ac} }}{{2a}}$

  ⇒             x = $\frac{{ - \left( { - 5} \right)\; \pm \;\sqrt {{{\left( { - 5} \right)}^2} - 4\left( 3 \right)\left( 2 \right)} }}{{2\left( 3 \right)}}$

 ⇒             x = $\frac{{5\; \pm \;\sqrt {25 - 24} }}{6}$

 ⇒           x = $\frac{{5 \pm 1}}{6}$ = $\frac{{5 + \;1}}{6}$ or $\frac{{5\; - \;1}}{6}$ = 1 or $\frac{2}{3}$ 

(ii) x² - 4x +5 = 0

Ans:  Discriminant = b² - 4ac = (-4)² - 4(1)(5) = 16-20 = -4 < 0 

Since D <0. Therefore, the equation has no real roots.

13.Solve : (i)$\frac{1}{{18 - x}}$-$\frac{1}{{18 + x}}$= $\frac{1}{{24}}$ and x > 0.

Ans: For $\frac{1}{{18 - x}}$-$\frac{1}{{18 + x}}$= $\frac{1}{{24}}$

⇒$\frac{{\left( {18 + x} \right) - \left( {18 - x} \right)}}{{\left( {18 + x} \right)\left( {18 - x} \right)}}$= $\frac{1}{{24}}$

⇒$\frac{{2x}}{{\left( {18 + x} \right)\left( {18 - x} \right)}}$= $\frac{1}{{24}}$

⇒ 24(2x)=(18+x)(18-x)

⇒ 48x = (324-x²)

⇒ x²+48x-324=0

⇒ x²+54x-6x-14=0

⇒ x(x+54)-6(x+54)=0

⇒ (x-6)(x+54)=0

Since, x-6=0 or x+54 = 0

⇒ x=6 or x=-54 

(ii) (x -10)$\left( {\frac{{1200}}{x} + 2} \right)$ = 1260 and x < 0.

Ans:  $\left( {\frac{{1200}}{x} + 2} \right)$ (x - 10) =1260

⇒ $\left( {\frac{{600 + x}}{x}} \right)$ (x - 10) = 630

⇒ (600 + x)(x - 10) = 630x

⇒ 600x - 6000 + x² -10x = 630x

⇒ x² - 40x - 6000 = 0

⇒ x² + 60x - 100x - 6000 = 0

⇒ x(x+60)-100(x+60) = 0

⇒ (x-100)(x+60) =0

Hence, x -100=0      or x+60 =0 

             x=100          or x = -60

But as x < 0, so x can't be positive. Hence, x=100

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