Question

Find the zeroes of the quadratic polynomial $p\left(x\right)=ab{x}^2+(b^2-ac)x-bc$ and verify the relationship between the zeroes and coefficients.

Answer 1

Step 1: To find the zeroes:

$\begin{array}{rcl}p\left(x\right)& =& ab{x}^2+(b^2-ac)x-bc\\ & =& ab{x}^2+b^{2}x-acx-bc\\ & =& bx(ax+b)-c(ax+b)\\ & =& (ax+b)(bx-c)\end{array}$

To find the zeroes of the polynomial, put $p\left(x\right)=0$

So, $(ax+b)(bx-c)=0$

$\begin{array}{rcl}& & \begin{array}{cc}\Rightarrow & ax+b=0\\ \Rightarrow & ax=-b\\ \Rightarrow & x=\frac{-b}{a}\end{array}\begin{array}{}\\ \\ \end{array}\end{array}$

or

$\begin{array}{l}\begin{array}{cc}\Rightarrow & bx-c=0\\ \Rightarrow & bx=c\\ \Rightarrow & x=\frac{c}{b}\end{array}\begin{array}{}\\ \\ \end{array}\end{array}$

Therefore, we have two zeroes, $x=-\frac{b}{a}\text{and}\frac{c}{b}$.

Step 2: To verify the relationship between the zeroes and coefficients

Let us take $\alpha =-\frac{b}{a}\text{and}\beta =-\frac{c}{b}$

$\text{Sum of the zeroes}=\alpha +\beta =-\frac{b}{a}+\frac{c}{d}=\frac{-b^2+ac}{ab}=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}=\frac{-b^2+ac}{ab}$$\text{Product of zeroes}=\alpha \beta =-\frac{b}{a}\times \frac{c}{b}=-\frac{bc}{ab}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}=-\frac{bc}{ab}$

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

Hence the zeroes of the quadratic polynomial are, $x=-\frac{b}{a},\frac{c}{b}$.