Question

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equations $2x^2+2(p+q)x+p^2+q^2=0$

Answer 1

We have,

$2x^2+2(p+q)x+(p^2+q^2)=0$

Let the roots of the required equation be $A$ and $B$.

Suppose that the roots of given equation be $a$ and $b$.

So,

Sum of roots$=-\frac{coeff.ofx}{coeff.of{x}^2}$

$a+b=-\frac{2(p+q)}{2}$

$a+b=-(p+q).......\left(1\right)$

Product of roots $=\frac{\text{constant}\text{term}}{coeff.of{x}^2}$

$\Rightarrow a.b=\frac{p^2+q^2}{2}......\left(2\right)$

From equation ()1 to and we get,

${(a+b)}^2={(p+q)}^2$

${(a-b)}^2={(a+b)}^2-4ab$

${(a-b)}^2={(p+q)}^2-4\frac{(p^2+q^2)}{2}$

${(a-b)}^2={(p+q)}^2-2(p^2+q^2)$

${(a-b)}^2=-{(p-q)}^2$

But according to the given question,

The values of square of sum of the roots and square of difference of the roots

Now,

$A+B=4pq$

$A.B={(p+q)}^2[-{(p-q)}^2]$

$A.B=-{(p^2-q^2)}^2$

The equation

$x^2-(A+B)x+A.B=0$

$\Rightarrow x^2-4pqx-{(p^2-q^2)}^2=0$

Hence, this is the answer.