NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.3

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;

I. $2{x^2} - 3x + 5 = 0$

II. $\quad 3{{\text{x}}^2} - 4\sqrt {3{\text{x}}}  + 4 = 0$

III. $\quad 2{{\text{x}}^2} - 6{\text{x}} + 3 = 0$

Ans: Given: Quadratic equations

To find: nature of the roots. 

We know that for a quadratic equation $a{x^2} + bx + c = 0$

Discriminant $ = {{\text{b}}^2} - 4{\text{ac}}$

(A) If ${b^2} - 4ac > 0 \to $ two distinct real roots

(B) If ${b^2} - 4ac = 0 \to $ two equal real roots

(C) If ${b^2} - 4ac < 0 \to $ no real roots

(I) $\quad 2{{\text{x}}^2} - 3{\text{x}} + 5 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} =  - 3,{\text{c}} = 5$

Discriminant : ${b^2} - 4ac$ $ = {( - 3)^2} - 4(2)(5) = 9 - 40$

$ =  - 31$

As ${b^2} - 4ac < 0$

Therefore, no real root is possible for the given equation.

(II) $3{{\text{x}}^2} - 4\sqrt {3{\text{x}}}  + 4 = 0$

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain $a = 3,b =  - 4\sqrt 3 ,c = 4$

Discriminant $:{b^2} - 4ac = {( - 4\sqrt 3 )^2} - 4(3)(4)$

$ = 48 - 48 = 0$

As ${b^2} - 4ac = 0$

Therefore, real roots exist for the given equation and they are equal to each other. And the roots will be $\dfrac{{ - b}}{{2a}}$ and $\dfrac{{ - b}}{{2a}}$.

$\dfrac{{ - b}}{{2a}} = \dfrac{{ - ( - 4\sqrt 3 )}}{{2 \times 3}} = \dfrac{{ - 4\sqrt 3 }}{6} = \dfrac{{2\sqrt 3 }}{3} = \dfrac{2}{{\sqrt 3 }}$

Therefore, the roots are $\dfrac{2}{{\sqrt 3 }}$ and $\dfrac{2}{{\sqrt 3 }}$

(III) $2{{\text{x}}^2} - 6{\text{x}} + 3 = 0$

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} =  - 6,{\text{c}} = 3$

Discriminant $ = {b^2} - 4ac = {( - 6)^2} - 4(2)(3)$

$ = 36 - 24 = 12$

As ${b^2} - 4ac > 0$

Therefore, distinct real roots exist for the given equation as follows. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$ = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)(3)} }}{{2(2)}}$

$ = \dfrac{{6 \pm \sqrt {12} }}{4} = \dfrac{{6 \pm 2\sqrt 3 }}{4}$

$ = \dfrac{{3 \pm \sqrt 3 }}{2}$

Therefore, the root are $\dfrac{{3 + \sqrt 3 }}{2}$ or $\dfrac{{3 - \sqrt 3 }}{2}$

2. Find the values of ${\text{k}}$ for each of the following quadratic equations, so that they have two equal roots.

I. $2{{\text{x}}^2} + {\text{kx}} + 3 = 0$

II. ${\text{kx}}({\text{x}} - 2) + 6 = 0$

Ans: Given: A quadratic equation

To find: k if the roots are equal 

We know that if an equation $a{x^2} + bx + c = 0$ has two equal roots, its Discriminant $\left( {{b^2} - 4ac} \right)$ will be 0 .

(I) $2{{\text{x}}^2} + {\text{kx}} + 3 = 0$

Comparing equation with $a{x^2} + bx + c = 0$, we obtain

${\text{a}} = 2,\;{\text{b}} = {\text{k}},{\text{c}} = 3$

Discriminant $ = {b^2} - 4ac = {(k)^2} - 4(2)(3)$

$ = {{\text{k}}^2} - 24$

For equal roots, Discriminant $ = 0$

${k^2} - 24 = 0$

${k^2} = 24$

$k =  \pm \sqrt {24}  =  \pm 2\sqrt 6 $

(II) ${\text{kx}}({\text{x}} - 2) + 6 = 0$

or $k{x^2} - 2kx + 6 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain $a = k,b =  - 2k,c = 6$

Discriminant $ = {b^2} - 4ac = {( - 2k)^2} - 4(k)(6)$

$ = 4{k^2} - 24k$

For equal roots, ${{\text{b}}^2} - 4{\text{ac}} = 0$

$4{{\text{k}}^2} - 24{\text{k}} = 0$

$4k(k - 6) = 0$

Either $4{\text{k}} = 0$ or ${\text{k}} = 6 = 0$

${\text{k}} = 0$ or ${\text{k}} = 6$

However, if ${\text{k}} = 0$, then the equation will not have the terms ' ${{\text{x}}^2}$ ' and ' ${\text{x}}$' .

Therefore, if the given equation has two equal roots, k should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\;{{\text{m}}^2}$ ? If so, find its length and breadth.

Ans: Given: A rectangular mango grove.

To find: its dimensions

Let us consider the breadth of mango grove be $l$. 

Length of mango grove will be 2l . Area of mango grove $ = (2l)(l) = 2{l^2}$ $ = 800$

${l^2} = \dfrac{{800}}{2} = 400$

${l^2} - 400 = 0$

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain $a = 1 ,b = 0,c = 400$

Discriminant $ = {b^2} - 4ac = {(0)^2} - 4 \times (1) \times ( - 400) = 1600$

Here, ${b^2} - 4ac > 0$

Therefore, the equation will have real roots.

$l = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$l = \dfrac{{ - 0 \pm \sqrt {1600}}}{{2*1}}$

And Therefore, the desired rectangular mango grove can be designed. $l =  \pm 20$

However, length cannot be negative. Therefore, breadth of mango grove $ = 20\;{\text{m}}$

Length of mango grove $ = 2 \times 20 = 40\;{\text{m}}$

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48 .

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not? 

Let us consider the age of one friend be $x$ years. 

Age of the other friend will be $(20 - {\text{x}})$ years.

4 years ago, age of 1 st friend $ = ({\text{x}} - 4)$ years 

And, age of ${2^{{\text{nd }}}}$ friend $ = (20 - {\text{x}} - 4)$ $ = (16 - {\text{x}})$ years

Given that, $(x - 4)(16 - x) = 48$

$16x - 64 - {x^2} + 4x = 48$

$ - {x^2} + 20x - 112 = 0$

${x^2} - 20x + 112 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 1,\;{\text{b}} =  - 20,{\text{c}} = 112$

Discriminant $ = {b^2} - 4ac = {( - 20)^2} - 4(1)(112)$

$ = 400 - 448 =  - 48$

As ${b^2} - 4ac < 0$

Therefore, no real root is possible for the given equation and Therefore, the given situation is not possible.

5. Is it possible to design a rectangular park of perimeter 80 and area $400\;{{\text{m}}^2}$ ? If so, find its length and breadth

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not? 

Let us consider the length and breadth of the park be l and ${\text{b}}$. 

Perimeter $ = 2(l + b) = 80$ $l + b = 40$

Or, ${\text{b}} = 40 - l$

Area $ = l \times {\text{b}} = l(40 - l) = 40l - {l^2}$

$40l - {l^2} = 400$

${l^2} - 40l + 400 = 0$

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain

${\text{a}} = 1,\;{\text{b}} =  - 40,{\text{c}} = 400$

Discriminate $ = {b^2} - 4ac = {( - 40)^2} - 4(1)(400)$

$ = 1600 - 1600 = 0$

As ${b^2} - 4ac = 0$

Therefore, the given equation has equal real roots. And Therefore, the given situation

is possible. Root of the given equation, $l =  - \dfrac{b}{{2a}}$

$l =  - \dfrac{{( - 40)}}{{2(1)}} = \dfrac{{40}}{2} = 20$

Therefore, length of park, $l = 20\;{\text{m}}$ 

And breadth of park, $b = 40 - l = 40 - 20 = 20\;{\text{m}}$

Conclusion

Class 10 Maths Exercise 4.3 Chapter 4 covers "Quadratic Equations." This exercise focuses on solving quadratic equations by completing the square, a key algebraic technique. Using NCERT Solutions by Vedantu enhances understanding of Quadratic Equations and the discriminant concept. These equations are vital in physics, engineering, economics, and more. Regular practice with Vedantu's NCERT solutions boosts comprehension and problem-solving skills. Focus on step-by-step solutions to grasp principles and ensure concept clarity.

Class 10 Maths Chapter 4: Exercises Breakdown

CBSE Class 10 Maths Chapter 4 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.