Question

Find zeroes of the following polynomials and verify the relation between the zeroes
(a) $6x^2+5x-1$
(b) $11x^2-8x-3$

Answer 1

The given polynomials are $6x^2+5x-1$ and $11x^2-8x-3$.

$\left(a\right)$ Let us first find the zeroes of the given polynomial $6x^2+5x-1$ by equating it to $0$ as shown below:

$6x^2+5x-1=0\Rightarrow 6x^2+6x-x-1=0\Rightarrow 6x(x+1)-1(x+1)=0\Rightarrow (6x-1)(x+1)=0\Rightarrow (6x-1)=0,(x+1)=0\Rightarrow 6x=1,x=-1\Rightarrow x=\frac{1}{6},x=-1$

Therefore, the zeroes are $\frac{1}{6}$ and $-1$.

In the polynomial $6x^2+5x-1$, we have $a=6,b=5$ and $-1$. Now, consider

Sum of zeroes

$\frac{1}{6}+(-1)=\frac{1}{6}-1=\frac{1-6}{6}=-\frac{5}{6}=-\frac{b}{a}$

Product of zeroes

$\frac{1}{6}\times (-1)=-\frac{1}{6}=\frac{c}{a}$

Hence, the relation between the zeroes and coefficients is verified.

$\left(b\right)$ Let us first find the zeroes of the given polynomial $11x^2-8x-3$ by equating it to $0$ as shown below:

$11x^2-8x-3=0\Rightarrow 11x^2-11x+3x-3=0\Rightarrow 11x(x-1)+3(x-1)=0\Rightarrow (11x+3)(x-1)=0\Rightarrow (11x+3)=0,(x-1)=0\Rightarrow 11x=-3,x=1\Rightarrow x=-\frac{3}{11},x=1$

Therefore, the zeroes are $-\frac{3}{11}$ and $1$.

In the polynomial $11x^2-8x-3$, we have $a=11,b=-8$ and $-3$. Now, consider

Sum of zeroes

$-\frac{3}{11}+1=\frac{-3+11}{11}=\frac{8}{11}=-\frac{(-8)}{11}=-\frac{b}{a}$

Product of zeroes

$-\frac{3}{11}\times 1=-\frac{3}{11}=\frac{c}{a}$

Hence, the relation between the zeroes and coefficients is verified.