First slide

Theory of equations

Question

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ If one of the roots is square of the other, then $p$ is equal to

Moderate

A

$\frac{1}{3}$
B

1
C

3
D

$2 / 3$

Solution

Let $α$ and $α^{2}$ be the roots of the given equation.
Then, $α + α^{2} = - p / 3$ and $α^{3} = 1$
Now, $α^{3} = 1 \Rightarrow α = 1, ω, ω^{2}$
Thus, the roots of the given equation are $ω$ and $ω^{2}$ Therefore
$α = 0$
Now,
$∴ α + α^{2} = - \frac{p}{3} \Rightarrow ω + ω^{2} = - \frac{p}{3} \Rightarrow - 1 = - \frac{p}{3} \Rightarrow p = 3$

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