Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients.
$x^2-2x-8$

Answer 1

Factorize the equation $x^2-2x-8$

Compare equation with $a{x}^2+bx+c=0$

We get, $a=1,b=-2,c=-8$

To factorize the value we have to find two value which

Sum is equal to, $b=-2$

product is $a\times c=1\times (-8)=-8$

So we can write middle term $=2x-4x$

We get, $x^2+2x-4x-8$

$\Rightarrow$ $x(x+2)-4(x+2)$

$\Rightarrow$ $(x+2)(x-4)$

Solve for first zero -

$x+2=0$

$\therefore$ $x=-2$

Solve for second zero -

$x-4=0$

$\therefore$ $x=4$

Sum of zero $-2+4=2$

product of zero $2\times (-4)=-8$

For equation $a{x}^2+bx+c=0$, if zero are $\alpha$ and $\beta$,

Plug the values of $a,b$ and $c$ we get

Sum of zeros $\frac{-b}{a}=-\frac{(-2)}{1}=2$

Product of zeros $\frac{c}{a}=\frac{-8}{1}=-8$

Hence we have verified that,

Sum of zeros $=\frac{-\left(Coefficientofx\right)}{Coefficientof{x}^2}$

Product of zeros $=\frac{Constantterm}{Coefficientof{x}^2}$