Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=x^2-x-4,$, find the value of $\frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta$

Answer 1

Consider the given equation.

$f\left(x\right)=x^2-x-4$

$\alpha$ and $\beta$ are the roots of this equation.

So,$\alpha +\beta =\frac{-b}{a}=-\frac{-1}{1}=1$

$\alpha \beta =\frac{c}{a}=\frac{-4}{1}=-4$

Since,$=\frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta$

$=\frac{\alpha +\beta }{\alpha \beta }-\alpha \beta$

Therefore,

$=\frac{1}{-4}-(-4)$

$=-\frac{1}{4}+4$

$=\frac{15}{4}$

Hence, the value is $\frac{15}{4}$.