Question

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

• A. $\frac{11}{5}$
• B. $\frac{14}{5}$
• C. $\frac{9}{5}$
• D. $\frac{7}{5}$

Answer 1

The correct option is B $\frac{14}{5}$

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

Here, $\alpha +\beta =\frac{-b}{a}=\frac{-4}{5}$ and $\alpha \beta =\frac{c}{a}=\frac{-9}{5}$

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

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

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

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

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

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

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

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

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

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