Question

$2ac = {b^2} + ab$ is the condition for the sum of the roots of $a{x^2} + bx + c = 0$ is equal to the sum of the squares of the root.
1) True
2) False

Answer 1

Hint:
The given equation is $a{x^2} + bx + c = 0$
We will find the real roots of the equation, by known methods. Either we can use the Sridharacharya formula or the factorisation method. Then equate the sum of roots and sum of squares of roots and finally, we will get our answer.

Complete step by step solution:
The given equation is $a{x^2} + bx + c = 0$
Let $\alpha $ and $\beta $ be the real roots of given quadratic equations
Sum of the roots, $\alpha + \beta = \dfrac{{ - b}}{a}$
Product of the roots, $\alpha \beta = \dfrac{c}{a}$
We have given that,
Sum of the roots = Sum of squares of the roots
i.e. $\dfrac{{ - b}}{a} = {a^2} + {b^2}$
\[\Rightarrow \dfrac{{ - b}}{a} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \]
$\Rightarrow \dfrac{{ - b}}{a} = {\left( {\dfrac{{ - b}}{a}} \right)^2} - \dfrac{{2c}}{a}$
$\Rightarrow - ab = {b^2} - 2ac$
$\Rightarrow ab + {b^2} = 2ac$

Hence the given question is true.

Note:
Since, the given equation is $a{x^2} + bx + c = 0$
Sum of the roots: The sum of the roots of the quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
Sum of the roots, $\alpha + \beta = \dfrac{{ - b}}{a}$
Product of the roots: The product of the roots of a quadratic equation is equal to the constant term
(the third term), divided by the leading coefficient.
Product of the roots, $\alpha \beta = \dfrac{c}{a}$