Question

Find the zeroes of the quadratic polynomial $2x^2-9-3x$ and verify the relationship between zeroes and the coefficients.

Answer 1

Step 1: Find the zeroes of quadratic polynomial $2x^2-9-3x$:

Let the polynomial $p\left(x\right)=2x^2-3x-9$.

$\begin{array}{rcl}p\left(x\right)& =& 2x^2-3x-9\\ & =& 2x^2-6x+3x-9\\ & =& 2x\left(x-3\right)+3\left(x-3\right)\\ & =& \left(x-3\right)\left(2x+3\right)\end{array}$

The value of polynomial $p\left(x\right)=0$.

$$\begin{gathered} \Rightarrow \left(x-3\right)\left(2x+3\right)=0 \\ \Rightarrow \left(x-3\right)=0or\left(2x+3\right)=0 \\ \Rightarrow x=3orx=\frac{-3}{2} \end{gathered}$$

Therefore, the zeroes of polynomial are $3\text{and}\frac{-3}{2}$.

Step 2. Verify the relationship between zeroes and coefficient:

Take $\alpha =3\text{and}\beta =\frac{-3}{2}$.

$\begin{array}{rcl}\text{sum of zeroes}& =& \alpha +\beta \\ & =& 3+\left(\frac{-3}{2}\right)\\ & =& \frac{3}{2}\\ & =& -\left(\frac{-3}{2}\right)\\ & =& -\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}\end{array}$

$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& 3\times \left(-\frac{3}{2}\right)\\ & =& \left(\frac{-9}{2}\right)\\ & =& \frac{\text{constant term}}{\text{coefficient of}{x}^2}\end{array}$

Hence, the zeroes are $3,\frac{-3}{2}$ and the relationship between the zeroes and the coefficients is verified.