Question

Find the zeros of quadratic polynomial $p\left(x\right)=4x^2+24x+36$ and verify the relationship between the zeros and their coefficients.

Answer 1

$P\left(x\right)=4x^2+24x+36$
Here, $a=4,b=24,c=36$
$p\left(x\right)=4x^2+12x+12x+36$
$=4x(x+3)+12(x+3)$
$=(x+3)(4x+12)$
$=4(x+3{)}^2$
So, the zeros of quadratic polynomial is $-3$ and $-3$.
Let $\alpha =-3$ and $\beta =-3$
Sum of zeros = $\alpha +\beta$$=-3-3$$=-6$

Sum of zeros $=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$
$-\frac{b}{a}=\frac{-24}{4}=-6$
So, sum of zeros $=\alpha +\beta =-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

Product of zeros $=\alpha \beta =(-3)(-3)=9$
Also, product of zeros $=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$
$=\frac{c}{a}=\frac{36}{4}=9$
So, product of zeros $=\alpha \beta =\frac{\text{constant term}}{\text{cofficient of}{x}^2}$