ISI2016-DCG-25

Question

If $\alpha$ and $\beta$ be the roots of the equation $x^{2}+3x+4=0,$ then the equation with roots $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$ is

• A. $x^{2}+2x+63=0$
• B. $x^{2}-63x+2=0$
• C. $x^{2}-2x-63=0$
• D. None of these

Answer 1

Answer: $\mathbf C$

Explanation:

For the given equation::

Let $\alpha$ and $\beta$ be the roots.

Now, sum of roots = $\alpha + \beta = \frac{-b}{a} = -3 \tag{1}$

product of roots = $\alpha\beta = \frac{c}{a} = 4 \tag{2}$

Now,

We need to find the equation having roots $(\alpha + \beta)^2$ and $(\alpha - \beta)^2$.

So, for this equation:

Sum of roots = $(\alpha + \beta)^2 \bf{+} (\alpha - \beta)^2 \implies 2(\alpha^2 + \beta^2)\implies 2[(\alpha + \beta)^2-2\alpha\beta]\tag{3}$

From equations, $\mathbf {1, 2, and\; 3}$, we get:

$\implies 2[(\alpha + \beta)^2-2\alpha\beta]\implies2[(-3)^2-2*4] = 2[9-8] = 2$

So, sum of roots coming out = $2$

Now, check the options and find the sum of their roots

For $A$ option, it is $-2$ and same for other options

For $C$ option, it is $2$

$\therefore \mathbf C$ is the correct option.