If  $α$  and  $β$  are the zeros of the quadratic polynomial  $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate: 
$a\left(\frac{α^2}{β}\ +\ \frac{β^2}{α}\right)\ +\ b\left(\frac{α}{β}\ +\ \frac{β}{α}\right)$

Given:

$α$  and  $β$  are the zeros of the quadratic polynomial  $f(x)\ =\ ax^2\ +\ bx\ +\ c$.

To do:

We have to find the value of $a\left(\frac{α^2}{β}\ +\ \frac{β^2}{α}\right)\ +\ b\left(\frac{α}{β}\ +\ \frac{β}{α}\right)$.

Solution:

The given quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.

Sum of the roots $= α+β = \frac{-b}{a}$.

Product of the roots $= αβ = \frac{c}{a}$.

We know that,

$ \begin{array}{l}
a\left(\frac{\alpha ^{2}}{\beta } +\frac{\beta ^{2}}{\alpha }\right) +b\left(\frac{\alpha }{\beta } +\frac{\beta }{\alpha }\right) =a\left(\frac{\alpha ^{3} +\beta ^{3}}{\alpha \beta }\right) +b\left(\frac{\alpha ^{2} +\beta ^{2}}{\alpha \beta }\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left( \alpha ^{3} +\beta ^{3}\right) +b\left( \alpha ^{2} +\beta ^{2}\right)}{\alpha \beta }\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left(( \alpha +\beta )^{3} -3\alpha \beta ( \alpha +\beta )\right) +b(\left(( \alpha +\beta )^{2} -2\alpha \beta \right)}{\alpha \beta }\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left(\left( -\frac{b}{a}\right)^{3} -3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right)\right) +b\left(\left(\frac{-b}{a}\right)^{2} -2\left(\frac{c}{a}\right)\right)}{\frac{c}{a}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left(\frac{-b^{3}}{a^{3}} +\frac{3bc}{a^{2}}\right) +b\left(\frac{b^{2}}{a^{2}} -\frac{2c}{a}\right)}{\frac{c}{a}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left(\frac{-b^{3} +3abc}{a^{3}}\right) +b\left(\frac{b^{2} -2ac}{a^{2}}\right)}{\frac{c}{a}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\frac{-b^{3} +3abc}{a^{2}} +\frac{b^{3} -2abc}{a^{2}}}{\frac{c}{a}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\frac{-b^{3} +3abc+b^{3} -2abc}{a^{2}}}{\frac{c}{a}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{abc}{ac}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =b
\end{array}$

The value of $a\left(\frac{α^2}{β}\ +\ \frac{β^2}{α}\right)\ +\ b\left(\frac{α}{β}\ +\ \frac{β}{α}\right)$ is $b$. 

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Updated on: 10-Oct-2022

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