Question

If \[a\] and \[b\] are the roots of the equation \[{x^2} + ax - b = 0\], then find \[a\] and \[b\].

Answer 1

Hint: To solve this we have to use the formula that is used to find the value of roots of a quadratic equation. the formula is \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] put the values of \[a,b,c\]. From here we get the values of roots and equate those roots to the given roots in question and find the value of the known. while taking out from the root check sign of that term also.

Complete answer:
Given,
A quadratic equation
\[{x^2} + ax - b = 0\]
To find,
The values of \[a\] and \[b\]if \[a,b\] are the roots of the quadratic equation \[{x^2} + ax - b = 0\]
To find the roots of this quadratic equation we have to use the formula that is used to find the roots of the general quadratic equation.\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Here, \[b\] is the coefficient of \[x\]
\[a\] is the coefficient of \[{x^2}\]
\[c\] is the constant term of the quadratic equation.
So in the given equation values of \[a,b,c\] are as follows.
\[ \Rightarrow a = 1\],
\[ \Rightarrow b = a\] and
\[ \Rightarrow c = - b\]
On putting all these values in the formula of finding the roots of the quadratic equation.
\[roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[roots = \dfrac{{ - a \pm \sqrt {{a^2} - 4\left( 1 \right)\left( { - b} \right)} }}{{2\left( 1 \right)}}\]
On further solving
\[roots = \dfrac{{ - a \pm \sqrt {{a^2} + 4b} }}{2}\]
First taking positive sign we get the first root of the equation.
\[root = \dfrac{{ - a + \sqrt {{a^2} + 4b} }}{2}\]
Now equate this with one of the root that is a
\[a = \dfrac{{ - a + \sqrt {{a^2} + 4b} }}{2}\]
On taking 2 on another side and taking \[a\] to the another side
\[2a + a = \sqrt {{a^2} + 4b} \]
On further solving
\[3a = \sqrt {{a^2} + 4b} \]
On squaring both side and \[{a^2}\] to another side
\[9{a^2} - {a^2} = 4b\]
On further solving
\[2{a^2} = b\] (i)
From here we get relation between \[a\] and \[b\]that is
\[ \Rightarrow 8{a^2} = 4b\]
now taking negative sign and comparing that equation by second root that is \[b\].
\[b = \dfrac{{ - a - \sqrt {{a^2} + 4b} }}{2}\]
On taking 2 on another side and taking \[a\] to the another side
\[2b + a = - \sqrt {{a^2} + 4b} \]
On squaring both side
\[{\left( {2b + a} \right)^2} = {a^2} + 4b\]
On further solving
\[4{b^2} + {a^2} + 4ab = {a^2} + 4b\]
\[4{b^2} + 4ab = 4b\]
Canceling 4 and b from both side
\[b + a = 1\] ……(ii)
On solving equation (i) and (ii)
On putting the value of on b from equation (i) to equation (ii)
\[2{a^2} + a = 1\]
On solving this equation we get two values of a that are
\[a = 0.5\] and
\[a = - 1\]
On putting these values in equation 2
\[b = 0.5\] and
\[b = 2\]
Final answer:
The values of \[a\] and \[b\] are
\[a = 0.5\] and \[b = 0.5\]
\[a = - 1\] and \[b = 2\]

Note:
To solve this type of question we have to use the formula that is used to find the roots of the quadratic equation. then assign the value of \[a,b,c\] according to the equation then put all those values in the formula of roots. You may commit a mistake in assigning the value of \[a,b,c\] and solving the root part of the formula.