Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $f\left(x\right)=5x^2+4x-9$ then evaluate the following: ${\alpha }^3-{\beta }^3$

• A. $\frac{874}{125}$
• B. $\frac{814}{125}$
• C. $\frac{894}{125}$
• D. $\frac{854}{125}$

Answer 1

Answer: (D)

$\frac{854}{125}$

$\alpha$ and $\beta$ are the zeros of the polynomial $f\left(x\right)=5x^2+4x-9$

Here,

$\alpha +\beta =\frac{-b}{a}=\frac{-4}{5}$

$\alpha \beta =\frac{c}{a}=\frac{-9}{5}$

Now,

${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow \frac{-4}{5}={\alpha }^2+{\beta }^2+2\frac{-9}{5}$

$\Rightarrow \frac{16}{25}={\alpha }^2+{\beta }^2+\frac{-18}{5}$

$\Rightarrow \frac{16}{25}+\frac{18}{5}={\alpha }^2+{\beta }^2$

$\Rightarrow \frac{16+90}{25}={\alpha }^2+{\beta }^2$

$\Rightarrow {\alpha }^2+{\beta }^2=\frac{106}{25}$

Again,

${(\alpha -\beta )}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

$\Rightarrow {(\alpha -\beta )}^2=\frac{106}{25}-2\frac{-9}{5}$

$\Rightarrow {(\alpha -\beta )}^2=\frac{106}{25}+\frac{18}{5}$

$\Rightarrow {(\alpha -\beta )}^2=\frac{106+90}{25}$

$\Rightarrow {(\alpha -\beta )}^2=\frac{196}{25}$

$\Rightarrow (\alpha -\beta )=\frac{14}{5}$

Now,

$({\alpha }^3-{\beta }^3)=(\alpha -\beta )({\alpha }^2+{\beta }^2+\alpha \beta )$

$\Rightarrow ({\alpha }^3-{\beta }^3)=\left(\frac{14}{5}\right)[\frac{106}{25}+\left(\frac{-9}{5}\right)]$

$\Rightarrow ({\alpha }^3-{\beta }^3)=\left(\frac{14}{5}\right)\left(\frac{106-45}{25}\right)$

$\Rightarrow ({\alpha }^3-{\beta }^3)=\left(\frac{14}{5}\right)\left(\frac{61}{25}\right)$

$\Rightarrow ({\alpha }^3-{\beta }^3)=\left(\frac{854}{125}\right)$