Question

If $\alpha$ and $\beta$ are the zeroes of $x^2-4x+1$, then find the value of $\frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta$.

• A. 3
• B. 5
• C. -5
• D. -3

Answer 1

The correct option is A 3

$x^2-4x+1=0$
Zeroes are $\alpha \&\beta$
$\frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta =?$
$\therefore \frac{\alpha +\beta }{\alpha \beta }-\alpha \beta =?$
For an equation $a{x}^2+bx+c=0$
Sum of zeros = $\frac{-b}{a}$
Product of zeros = $\frac{c}{a}$
$\therefore \alpha +\beta =4\&\alpha \beta =1$
$\therefore \frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta =\frac{\alpha +\beta }{\alpha \beta }-\alpha \beta$
$=\frac{4}{1}-1$
$=4-1$
$\therefore \frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta =3$