Show that $\frac{1}{2}$ and $\frac{- 3}{2}$ are the zeroes of the polynomial $4 x^{2} + 4 x - 3$ and also verify the relationship between the zeroes and the coefficients of the polynomial.

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Solution

$4 x^{2} + 4 x - 3$ splitting middle terms

$4 x^{2} - 2 x + 6 x - 3 = 0$

$\Rightarrow 2 x (2 x - 1) + 3 (2 x - 1) = 0$

$\Rightarrow (2 x + 3) (2 x - 1) = 0$

So $x = \frac{- 3}{2}$ & $x = \frac{1}{2}$

So $(- \frac{3}{2}, \frac{1}{2})$ are the roots of the given equation.

$α + β = \frac{- b}{a}$

$\Rightarrow \frac{- 3}{2} + \frac{1}{2} = \frac{- 2}{2} = - 1$

$\frac{- b}{a} = \frac{- 4}{4} = - 1$

$α β = (\frac{- 3}{2}) (\frac{1}{2}) = \frac{- 3}{4} = \frac{c}{a}$

So $α + β = \frac{- b}{a}$ & $α β = \frac{c}{a}$
Hence proved.

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