Question

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

Answer 1

Step 1: Form the equation

The given polynomial is $6x^2-3-7x$.

We know that the zeroes of a polynomial are evaluated by equating them with zero.

$\therefore p\left(x\right)=0$

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

Step 2: Solve to find the zeroes

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

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

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

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

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

Step 3: Verification

We know that for a given polynomial $a{x}^2+bx+c$,

Sum of the zeroes$=-\frac{b}{a}$ and product of the roots$=\frac{c}{a}$

Sum of the zeroes $=\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{3}$

Again, $-\frac{b}{a}=-\frac{\left(-7\right)}{6}=\frac{7}{3}$

Product of the zeroes $=\frac{3}{2}\times \left(-\frac{1}{3}\right)=-\frac{3}{6}=-\frac{1}{2}$

Again, $\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}$

Thus, the relationship between the zeroes and the coefficients of the polynomial is verified.

Hence, the zeroes of this polynomial are $\frac{3}{2}$ and $-\frac{1}{3}$.