Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients.
$4s^2-4s+1$

Answer 1

Factorize the equation $4s^2-4s+1=0$

Compare equation with $a{s}^2+bs+c=0$

We get, $a=4,b=-4,c=1$

To factorize the value we have to find two value which

Sum is equal to, $b=-4$

product is $a\times c=4\times \left(1\right)=4$

$-2$ and $-2$ are required values which

sum is $(-2)+(-2)=-4$

product is $(-2)\times (-2)=4$

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

We get, $4s^2-2s-2s+1=0$

$\Rightarrow$ $2s(2s-1)-1(2s-1)=0$

$\Rightarrow$ $(2s-1)(2s-1)=0$

Solve for first zero -

$2s-1=0$

$\therefore$ $s=\frac{1}{2}$

Solve for second zero -

$2s-1=0$

$\therefore$ $s=\frac{1}{2}$

Sum of zero $\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$

product of zero $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

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

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

Sum of zeros $\frac{-b}{a}=-\frac{(-4)}{4}=1$

Product of zeros $\frac{c}{a}=\frac{1}{4}=\frac{1}{4}$

Hence we have verified that,

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

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