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If sum of the roots of $3 x^{2}+(3 p+1) x-(p+5)=0$ is equal to their product, then what is the value of $p ?$
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Verified Answer
The correct answer is:
2
Let $\alpha$ and $\beta$ are the roots of the given equation.
$\therefore$
$\alpha+\beta=\frac{-(3 p+1)}{3}$
and $\quad \alpha \beta=\frac{-(p+5)}{3}$
Now, $\frac{-(3 p+1)}{3}=\frac{-(p+5)}{3}$
$\Rightarrow 3 p+1=p+5$
$\therefore \quad 2 p=4$
$\therefore \quad p=2$
$\therefore$
$\alpha+\beta=\frac{-(3 p+1)}{3}$
and $\quad \alpha \beta=\frac{-(p+5)}{3}$
Now, $\frac{-(3 p+1)}{3}=\frac{-(p+5)}{3}$
$\Rightarrow 3 p+1=p+5$
$\therefore \quad 2 p=4$
$\therefore \quad p=2$
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