Question

Find the zeroes of the quadratic polynomial $2x^2-3+5x$ and verify the relationship between zeroes and the coefficients.

Answer 1

Step 1: Find the zeroes of quadratic polynomial $2x^2-3+5x$:

Let the polynomial $p\left(x\right)=2x^2+5x-3$.

$\begin{array}{rcl}p\left(x\right)& =& 2x^2+5x-3\\ & =& 2x^2+6x-x-3\\ & =& 2x\left(x+3\right)-1\left(x+3\right)\\ & =& \left(2x-1\right)\left(x+3\right)\end{array}$

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

$$\begin{gathered} \Rightarrow \left(2x-1\right)\left(x+3\right)=0 \\ \Rightarrow \left(2x-1\right)=0or\left(x+3\right)=0 \\ \Rightarrow x=\frac{1}{2}orx=-3 \end{gathered}$$

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

Step 2. Verify the relationship between zeroes and coefficient:

Take $\alpha =\frac{1}{2}\text{and}\beta =-3$.

$\begin{array}{rcl}\text{sum of zeroes}& =& \alpha +\beta \\ & =& \frac{1}{2}-3\\ & =& \frac{-5}{2}\\ & =& -\left(\frac{5}{2}\right)\\ & =& -\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}\end{array}$

$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& \frac{1}{2}\times \left(-3\right)\\ & =& \left(\frac{-3}{2}\right)\\ & =& \frac{\text{constant term}}{\text{coefficient of}{x}^2}\end{array}$

Hence, the zeroes are $-3,\frac{1}{2}$ and the relationship between the zeroes and the coefficients is verified.