Question

Find the zeroes of the quadratic polynomial $4x^2-4x-3$ and verify the relation between the zeroes and their coefficients.

Answer 1

Step 1: Find the zeroes of quadratic polynomial $4x^2-4x-3$:

Let the polynomial $p\left(x\right)=4x^2-4x-3$.

$\begin{array}{rcl}p\left(x\right)& =& 4x^2-4x-3\\ & =& 4x^2-6x+2x-3\\ & =& 2x\left(2x-3\right)+1\left(2x-3\right)\\ & =& \left(2x-3\right)\left(2x+1\right)\end{array}$

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

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

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

Step 2. Verify the relationship between zeroes and coefficient:

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

$\begin{array}{rcl}\text{sum of zeroes}& =& \alpha +\beta \\ & =& \frac{3}{2}-\frac{1}{2}\\ & =& \frac{2}{2}\\ & =& -\left(\frac{-4}{4}\right)\\ & =& -\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}\end{array}$

$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& \frac{3}{2}\times \left(-\frac{1}{2}\right)\\ & =& \left(\frac{-3}{4}\right)\\ & =& \frac{\text{constant term}}{\text{coefficient of}{x}^2}\end{array}$

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