If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate:
$α^4\ +\ β^4$

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 $α^4\ +\ β^4$.

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}
\alpha ^{4} +\beta ^{4} =\left( \alpha ^{2}\right)^{2} +\left( \beta ^{2}\right)^{2}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\left( \alpha ^{2} +\beta ^{2}\right)^{2} -2\alpha ^{2} \beta ^{2}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\left(( \alpha +\beta )^{2} -2\alpha \beta \right)^{2} -2( \alpha \beta )^{2}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\left(\left( -\frac{b}{a}\right)^{2} -2\left(\frac{c}{a}\right)\right)^{2} -2\left(\frac{c}{a}\right)^{2}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\left(\frac{b^{2}}{a^{2}} -\frac{2c}{a}\right)^{2} -\left(\frac{2c^{2}}{a^{2}}\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\left(\frac{b^{2} -2ac}{a^{2}}\right)^{2} -\left(\frac{2c^{2}}{a^{2}}\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( b^{2} -2ac\right)^{2} -2a^{2} c^{2}}{a^{4}}
\end{array}$

The value of $α^4\ +\ β^4$ is $\frac{(b^2-2ac)^2-2a^2c^2}{a^4}$.

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Updated on: 10-Oct-2022

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