Question

Form the quadratic equation whose roots are the squares of the sum of the roots and square of the difference of the roots of the equation $2{{x}^{2}}+2\left( m+n \right)x+{{m}^{2}}+{{n}^{2}}=0$?

Answer 1

Hint:We first express the general information about the characteristics of the roots of a quadratic equation. We get that for quadratic equation $d{{x}^{2}}+ex+f=0$, if the roots are $p,q$ then we can say that $p+q=-\dfrac{e}{d}$ and $pq=\dfrac{f}{d}$. We put the given values of the roots in the equation. Then we find the squares of the sum of the roots and square of the difference of the roots to find new roots for the new equation. We can get the equation by taking the form of ${{x}^{2}}-\left( p+q \right)x+pq=0$.

Complete step by step answer:
We know that for quadratic equation $d{{x}^{2}}+ex+f=0$, if the roots are $p,q$ then we can say that $p+q=-\dfrac{e}{d}$ and $pq=\dfrac{f}{d}$. Let’s assume the roots of the equation $2{{x}^{2}}+2\left( m+n \right)x+{{m}^{2}}+{{n}^{2}}=0$ are $p,q$. We get $p+q=-\dfrac{2\left( m+n \right)}{2}=m+n$ and $pq=\dfrac{{{m}^{2}}+{{n}^{2}}}{2}$.Now we find the squares of the sum of the roots and the square of the difference of the roots.
${{\left( p+q \right)}^{2}}={{\left( m+n \right)}^{2}}$
$\Rightarrow {{\left( p-q \right)}^{2}}={{\left( p+q \right)}^{2}}-4pq \\
\Rightarrow {{\left( p-q \right)}^{2}}={{\left( m+n \right)}^{2}}-2\left( {{m}^{2}}+{{n}^{2}} \right)=-{{\left( m-n \right)}^{2}} \\ $
Therefore, ${{\left( m+n \right)}^{2}}$ and $-{{\left( m-n \right)}^{2}}$ are the new roots of the new equation. We form the new equation as ${{x}^{2}}-\left( p+q \right)x+pq=0$ for roots $p,q$.
${{x}^{2}}-\left\{ {{\left( m+n \right)}^{2}}-{{\left( m-n \right)}^{2}} \right\}x-{{\left( m+n \right)}^{2}}{{\left( m-n \right)}^{2}}=0 \\
\therefore {{x}^{2}}-4mnx-{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=0 $

Therefore, the new quadratic equation is ${{x}^{2}}-4mnx-{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=0$.

Note: We can also put the roots in the equation to find if the roots satisfy the equation which means the final answer will be 0. We find the value of the root for which the equation ${{x}^{2}}-4mnx-{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=0$ is satisfied. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We express it as $\left\{ x-{{\left( m+n \right)}^{2}} \right\}\left\{ x+-{{\left( m-n \right)}^{2}} \right\}=0$.