# Form the quadratic equation whose roots are the squares of the sum of the roots and square of the difference of the root of the equation 2x^2+2(m+n)x+m^2+n^2=0.?

Jun 1, 2016

Desired equation is ${x}^{2} - 4 m n x - {\left({m}^{2} - {n}^{2}\right)}^{2} = 0$

#### Explanation:

Let $\alpha$ and $\beta$ be the roots of the equation $2 {x}^{2} + 2 \left(m + n\right) x + {m}^{2} + {n}^{2} = 0$

As such $\alpha + \beta = - 2 \frac{m + n}{2} = - \left(m + n\right)$

and $\alpha \times \beta = \frac{{m}^{2} + {n}^{2}}{2}$

We have to find the equation whose roots are ${\left(\alpha + \beta\right)}^{2}$ and ${\left(\alpha - \beta\right)}^{2}$.

Sum of these roots will be ${\left(\alpha + \beta\right)}^{2} + {\left(\alpha - \beta\right)}^{2} = 2 \left({\alpha}^{2} + {\beta}^{2}\right)$

= $2 \left({\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta\right) = 2 \left({\left(- \left(m + n\right)\right)}^{2} - 2 \frac{{m}^{2} + {n}^{2}}{2}\right)$

= $2 {\left(m + n\right)}^{2} - 2 \left({m}^{2} + {n}^{2}\right) = 4 m n$

Product of these roots will be ${\left(\alpha + \beta\right)}^{2} \times {\left(\alpha - \beta\right)}^{2}$

= ${\left({\alpha}^{2} - {\beta}^{2}\right)}^{2} = {\left({\alpha}^{2} + {\beta}^{2}\right)}^{2} - 4 {\alpha}^{2} {\beta}^{2}$

= ${\left({\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta\right)}^{2} - 4 {\alpha}^{2} {\beta}^{2}$

= ${\left({\left(- m - n\right)}^{2} - 2 \frac{{m}^{2} + {n}^{2}}{2}\right)}^{2} - 4 {\left(\frac{{m}^{2} + {n}^{2}}{2}\right)}^{2}$

= $4 {m}^{2} {n}^{2} - {m}^{4} - {n}^{4} - 2 {m}^{2} {n}^{2} = - {m}^{4} - {n}^{4} + 2 {m}^{2} {n}^{2}$

= $- {\left({m}^{2} - {n}^{2}\right)}^{2}$

Hence equation will be

${x}^{2} -$(sum of roots)$x +$product of roots$= 0$ or

${x}^{2} - 4 m n x - {\left({m}^{2} - {n}^{2}\right)}^{2} = 0$