Question

Find the zeros of the quadratic polynomial $f\left(x\right)=x^2-2x-8$ and verify the relationship between the zeros and their coefficients.

Answer 1

$f\left(x\right)=x^2-2x-8$
$\Rightarrow f\left(x\right)=x^2-4x+2x-8$
$\Rightarrow f\left(x\right)=x(x-4)+2(x-4)]$
$\Rightarrow f\left(x\right)=(x-4)(x+2)$
Zeros of f(x) are given by f(x) = 0
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$
$\Rightarrow x=4$ or $x=-2$
So, $\alpha =4$ and $\beta =-2$
$\therefore$ sum of zeros $=\alpha +\beta =4-2=2$
Also, sum of zeros $=\frac{\text{Coefficient of x}}{\text{Coefficient of}{x}^2}$
$=\frac{-(-2)}{1}=2$
So, sum of zeros $=\alpha +\beta =-\frac{\text{Coefficient of x}}{\text{Coefficient of}{x}^2}$
Now, product of zeros $=\alpha \beta =\left(4\right)(-2)=-8$
Also, product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of}{x}^2}$
$=\frac{-8}{1}=-8$
$\therefore$ Product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of}{x}^2}=\alpha \beta$