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

Given: The polynomial $4x^2+4x-3$.

To do: To show that $\frac{1}{2}$ and $-\frac{3}{2}$ are the zeroes of the polynomial $4x^2+4x-3$.

Solution:

Let $p( x)=4x^2+4x-3$

If $\frac{1}{2}$ and $-\frac{3}{2}$ are the zeroes of the polynomial $4x^2+4x-3$, then they will satisfy the polynomial.

$\Rightarrow p( \frac{1}{2})=4( \frac{1}{2})^2+4( \frac{1}{2})-3$

$\Rightarrow p( \frac{1}{2})=4( \frac{1}{4})+2-3$

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

$\Rightarrow p( \frac{1}{2})=0$

Therefore, $\frac{1}{2}$ is the zeroes of the polynomial $4x^2+4x-3$.

Now, $p( -\frac{3}{2})=4( -\frac{3}{2})^2+4( -\frac{3}{2})-3$

$\Rightarrow p( -\frac{3}{2})=4( \frac{9}{4})+2( -3)-3$

$\Rightarrow p( -\frac{3}{2})=9-6-3$

$\Rightarrow p( -\frac{3}{2})=0$

Thus, it has been proved that $\frac{1}{2}$ and $-\frac{3}{2}$ are the zeroes of the polynomial $4x^2+4x-3$

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Updated on: 10-Oct-2022

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