Question

The roots of the equation $9{{x}^{2}}-6{{b}^{2}}x-\left( {{a}^{4}}-{{b}^{4}} \right)=0$ are $\dfrac{{{a}^{2}}+{{b}^{2}}}{3},\dfrac{{{b}^{2}}-{{a}^{2}}}{3}$. Is it true or false?

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 $m,n$ then we can say that $m+n=-\dfrac{e}{d}$ and $mn=\dfrac{f}{d}$. We put the given values of the roots in the equation to find if the statement is true or false.

Complete step by step solution:
We know that for quadratic equation $d{{x}^{2}}+ex+f=0$, if the roots are $m,n$ then we can say that
$m+n=-\dfrac{e}{d}$ and $mn=\dfrac{f}{d}$.
Let’s assume the roots of the equation $9{{x}^{2}}-6{{b}^{2}}x-\left( {{a}^{4}}-{{b}^{4}} \right)=0$ is $m=\dfrac{{{a}^{2}}+{{b}^{2}}}{3},n=\dfrac{{{b}^{2}}-{{a}^{2}}}{3}$.
We now find the addition and multiplication of the roots and get
$m+n=\dfrac{{{a}^{2}}+{{b}^{2}}}{3}+\dfrac{{{b}^{2}}-{{a}^{2}}}{3}=\dfrac{{{a}^{2}}+{{b}^{2}}+{{b}^{2}}-{{a}^{2}}}{3}=\dfrac{2{{b}^{2}}}{3}$.
We know the formula of \[\left( p+q \right)\left( p-q \right)={{p}^{2}}-{{q}^{2}}\]
The multiplication gives \[mn=\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{3} \right)\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{3} \right)=\dfrac{{{\left( {{b}^{2}} \right)}^{2}}-{{\left( {{a}^{2}} \right)}^{2}}}{9}=\dfrac{{{b}^{4}}-{{a}^{4}}}{9}\].
We take the equation $9{{x}^{2}}-6{{b}^{2}}x-\left( {{a}^{4}}-{{b}^{4}} \right)=0$ and equate with $d{{x}^{2}}+ex+f=0$ to find the value of $-\dfrac{e}{d}$ and $\dfrac{f}{d}$. We have $d=9,e=-6{{b}^{2}},f=-\left( {{a}^{4}}-{{b}^{4}} \right)$
We get \[-\dfrac{e}{d}=-\left( -\dfrac{2{{b}^{2}}}{3} \right)=\dfrac{2{{b}^{2}}}{3}=m+n\] and $\dfrac{f}{d}=-\dfrac{{{a}^{4}}-{{b}^{4}}}{9}=\dfrac{{{b}^{4}}-{{a}^{4}}}{9}=mn$.
Thus, we can show that the roots $\dfrac{{{a}^{2}}+{{b}^{2}}}{3},\dfrac{{{b}^{2}}-{{a}^{2}}}{3}$ satisfy the equation $9{{x}^{2}}-6{{b}^{2}}x-\left( {{a}^{4}}-{{b}^{4}} \right)=0$. The given statement is true.

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 $9{{x}^{2}}-6{{b}^{2}}x-\left( {{a}^{4}}-{{b}^{4}} \right)$ is satisfied. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.