Question

If $\alpha$ and $\beta$ be the roots of the quadratic equation $2 x^{2}+2(a+b) x+a^{2}+b^{2}=0$, then the equation whose roots are $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$ is
A. $x^{2}-2 a b x-\left(a^{2}-b^{2}\right)^{2}=0$
B. $x^{2}-4 a b x-\left(a^{2}-b^{2}\right)^{2}=0$
C. $x^{2}-4 a b x+\left(a^{2}-b^{2}\right)^{2}=0$
D. None of these

Answer 1

Hint: We are given an equation and its roots and we are asked to find an equation whose roots are $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$. To solve this question, we need to recall the relationship between the roots and the coefficients of a polynomial. Also, recollect given the roots of the polynomial how to build the equation.

Formula Used: Let $a x^{2}+b x+c$ be a polynomial having $\alpha$ and $\beta$ as the zeros of the polynomial. Then,
Sum of the roots of the polynomial, $\alpha+\beta= -\dfrac{\text { coefficient of } x}{\text { coefficient of } x^{2}}= -\dfrac{b}{a}$
Product of the roots of the polynomial, $\alpha \beta=\dfrac{\text { constant term }}{\text { coefficient of } x^{2}}=\dfrac{c}{a}$

Complete step by step solution: We have the quadratic equation, $2 x^{2}+2(a+b) x+a^{2}+b^{2}=0$ and $\alpha$ and $\beta$ are its roots.
Sum of the roots, $\alpha+\beta=-\dfrac{2(a+b)}{2}= -(a+b)$ ------- (1)
Product of the roots, $\alpha \beta=\dfrac{a^{2}+b^{2}}{2}$ ------- (2)
Now, we know that if the roots are given we can form the quadratic equation.
That is, the quadratic equation can be written as
$x^{2}-(\text { sum of the roots }) x+\text { product of the roots }=0 $
Here, we have the roots $-(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$
$(\alpha+\beta)^{2}=(-(a+b))^{2}=(a+b)^{2}$ ------ (3)
$(\alpha-\beta)^{2}=\alpha^{2}+\beta^{2}-2 \alpha \beta$ ------ (4)
Now, we have
$(\alpha+\beta)^{2}=(a+b)^{2}$
$\alpha^{2}+\beta^{2}+2 \alpha \beta=a^{2}+b^{2}+2ab$
$\alpha^{2}+\beta^{2}=a^{2}+b^{2}+2 a b-2 \alpha \beta$
Substitute, $\alpha \beta=\dfrac{a^{2}+b^{2}}{2}$ in the above equation we get,
$\alpha^{2}+\beta^{2}=a^{2}+b^{2}+2 a b-2 \times \dfrac{a^{2}+b^{2}}{2}$
$\alpha^{2}+\beta^{2}=a^{2}+b^{2}+2 a b-a^{2}-b^{2}$
$\alpha^{2}+\beta^{2}=2ab$ ------- (5)
Substitute equation (2) and equation (5) in equation (4) and we get
$(\alpha-\beta)^{2}=2 a b-2 \times \dfrac{a^{2}+b^{2}}{2}$
$(\alpha-\beta)^{2}=2 a b-\left(a^{2}+b^{2}\right)$
$(\alpha-\beta)^{2}=-\left(a^{2}+b^{2}-2 a b\right)$
$(\alpha-\beta)^{2}=-(a-b)^{2}$ ------ (6)
Now, we form the required quadratic equation,
$x^{2}-\left[(\alpha+\beta)^{2}+(\alpha-\beta)^{2}\right] x+\left[(\alpha+\beta)^{2} \times(\alpha-\beta)^{2}\right]$
Substitute equation (3) and equation (6) in the above equation we get,
$x^{2}-\left[(a+b)^{2}-(a-b)^{2}\right] x+\left[(a+b)^{2} \times-(a-b)^{2}\right]$
$x^{2}-\left[a^{2}+b^{2}+2 a b-\left(a^{2}+b^{2}-2 a b\right)\right] x-\left[(a+b)^{2}(a-b)^{2}\right]$
$x^{2}-\left[a^{2}+b^{2}+2 a b-a^{2}-b^{2}+2 a b\right]-[(a+b)(a-b)(a+b)(a-b)]$
$x^{2}-[2 a b+2 a b] x-\left[\left(a^{2}-b^{2}\right)\left(a^{2}-b^{2}\right)\right]$
$x^{2}-4 a b x-\left(a^{2}-b^{2}\right)^{2}$
Therefore, the required quadratic equation is $x^{2}-4 a b x-\left(a^{2}-b^{2}\right)^{2}$.

So, Option ‘B’ is correct

Note: Be careful with the substitution part, do not go with the unnecessary expansion of brackets instead look for ways in which we can make the expansion part simpler. Also, look for the signs in the equation that we used for forming the required equation.