Question

Find the zeroes of the quadratic polynomial f(x) = 4x² - 4x - 3 and verify the relation between its zeroes and coefficients.

Answer 1

We have:

f(x) = 4x² - 4x - 3

= 4x² ˗ (6x ˗ 2x) ˗ 3

= 4x² ˗ 6x + 2x ˗ 3

= 2x (2x ˗ 3) + 1(2x ˗ 3)

= (2x + 1) (2x ˗ 3)

∴ f(x) = 0 ⇒ (2x + 1) (2x ˗ 3) = 0

⇒ 2x + 1= 0 or 2x ˗ 3 = 0`

⇒ ${\displaystyle x=\frac{-1}{2}\text{or}x=\frac{3}{2}}$ 

So, the zeroes of f(x) are ${\displaystyle \frac{-1}{2}\text{and}\frac{3}{2}}$ 

Sum of zeroes = ${\displaystyle (-\frac{1}{2})+\left(\frac{3}{2}\right)=\frac{-1+3}{2}=\frac{2}{2}=1=\frac{\text{-(Coefficient of x)}}{(\text{Coefficient of} x^2)}}$ 

Product of zeroes =${\displaystyle \left(\frac{-1}{2}\right)\times \left(\frac{3}{2}\right)=\frac{-3}{4}=\frac{\text{Constant term}}{(\text{Coefficient of} x^2)}}$