Question

If m and n are the roots of the equation ${{x}^{2}}-6x+2=0$ find the value of ${{m}^{3}}{{n}^{2}}+{{n}^{3}}{{m}^{2}}$ .

Answer 1

Hint: A quadratic equation is an equation of the form $a{{x}^{2}}+bx+c=0$. It has two roots which can be real or imaginary and repeated or distinct. So, let roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ be $\alpha $ and$\beta $ .Then, we can write the quadratic equation $a{{x}^{2}}+bx+c=0$ in the form $a(x-\alpha )(x-\beta )=0$.

Complete step-by-step answer:
So, $a{{x}^{2}}+bx+c=a(x-\alpha )\left( x-\beta \right)$
So, after rearranging them $a\left( x-\alpha \right)\left( x-\beta \right)=a{{x}^{2}}+bx+c$
We get as follows ,
                  \[a\left( {{x}^{2}}-\beta x-\alpha x+\alpha \beta \right)=a{{x}^{2}}+bx+c\]
            $\Rightarrow a\left( {{x}^{2}}-\left( \beta +\alpha \right)x+\alpha \beta \right)=a({{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a})$
If $a\ne 0$( that is the equation is quadratic ), then, ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta ={{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}$
Now, we compare both sides of equation we get that, $\alpha +\beta =-\dfrac{b}{a}$ and also, $\alpha \beta =\dfrac{c}{a}$
So, if the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ be$\alpha $ and $\beta $. Then, the sum of roots is equal to $\left( -\dfrac{b}{a} \right)$ and product of roots is equal to $\left( \dfrac{c}{a} \right)$. Now, we use this relation in the quadratic equation to find the required value.

Now, the given quadratic equation is
 ${{x}^{2}}-6x+2=0$
Let the roots be $m$ and $n$ of the given quadratic equation
So, using the formula for quadratic equation, we get;
 $a=1$, $b=-6$ and $c=2$
So, using the formula the for sum of roots and product of roots, we get;
 $m+n=-\dfrac{b}{a}=-\left( -\dfrac{6}{1} \right)=6$
 $mn=\dfrac{c}{a}=\dfrac{2}{1}=2$
Now, given in the question, to find ${{m}^{3}}{{n}^{2}}+{{n}^{3}}{{m}^{2}}$which is equal to ${{m}^{2}}{{n}^{2}}\left( m+n \right)$
Now, by putting values of $m+n$ and $mn$ , we get;
 ${{\left( mn \right)}^{2}}\left( m+n \right)={{2}^{2}}\times \left( 6 \right)=24$

Note: In this question one may think to find out first $m$ and $n$ . And then put values in ${{m}^{3}}{{n}^{2}}+{{n}^{3}}{{m}^{2}}$ but that will lead to unnecessary cumbersome calculation, first of finding $m$ and $n$ and then finding values of ${{m}^{3}}$ , ${{n}^{3}}$, ${{m}^{2}}$ ,${{n}^{2}}$ which should be avoided.