Find the zeroes of $P( x)=2x^{2}-x-6$ and verify the relation of zeroes with these coefficients.
Given: $P( x)=2x^{2}-x-6$
To do: To find the zeroes of $P( x)=2x^{2}-x-6$ and verify the relation of zeroes with these coefficients.
Solution:
$2x^{2}-x -6$
by splitting the middle term,
$2x^{2}-4x+3x-6$
$\Rightarrow 2x(x-2) + 3(x-2)$
$\Rightarrow (2x+3) (x-2) = 0$
$\Rightarrow 2x+3=0$ & $x-2=0$
If $2x+3=0$
$\Rightarrow x= -3/2$
And if $x-2=0$
$Rightarrow x=2$
Verification:
Here $a=2$, $b=-1$ & $c=-6$
$\alpha+\beta= \frac{-b}{a} =\frac{1}{2}$
$\alpha\beta=\frac{c}{a}=\frac{-6}{2}=-3$
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