Question

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
$6x^2-3-7x$

Answer 1

Let $f\left(x\right)=6x^2-3-7x$

To calculate the zeros of the given equation, put $f\left(x\right)=0$.

$6x^2-7x-3=0$

$6x^2-9x+2x-3=0$

$3x\left(2x-3\right)+1\left(2x-3\right)=0$

$\left(3x+1\right)\left(2x-3\right)=0$

$x=-\frac{1}{3},x=\frac{3}{2}$

The zeros of the given equation is $-\frac{1}{3}$ and $\frac{3}{2}$.

Sum of the zeros is $-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}$.

Product of the zeros is $-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}$.

According to the given equation

The sum of the zeros is,

$\frac{-b}{a}=\frac{-\left(-7\right)}{6}$

$=\frac{7}{6}$

The product of the zeros is

$\frac{c}{a}=\frac{-3}{6}$

$=-\frac{1}{2}$

Hence, it is verified that,

$sumofzeros=\frac{-coefficientofx}{coefficientof{x}^2}$

And,

$productofzeros=\frac{constantterm}{coefficientof{x}^2}$