Question

For the equation $3x^2+px+3=0,p>0,$ if one of the roots is the square of the other, then find the value of $p.$

• A. $\frac{1}{3}$
• B. $1$
• C. $3$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$3$

Given the quadratic equation:
$3x^2+px+3=0,p>0,$
Now, let $\alpha$ be one of the roots.

$\Rightarrow \alpha +{\alpha }^2=-\frac{p}{3}\cdots \left(i\right)\&\alpha \cdot {\alpha }^2=\frac{3}{3}=1\Rightarrow {\alpha }^3=1\cdots \left(ii\right)$
Now, ${\alpha }^3=1$
$\Rightarrow (\alpha -1)({\alpha }^2+\alpha +1)=0\Rightarrow (\alpha -1)=0\text{or}({\alpha }^2+\alpha +1)=0$
Now, let $\alpha =1$
$\Rightarrow 1+1=\frac{-p}{3}\Rightarrow p=-6$
Which is rejected as $p>0$

Now, let $({\alpha }^2+\alpha +1)=0\Rightarrow \alpha +{\alpha }^2=-1$
$\Rightarrow \alpha +{\alpha }^2=-\frac{p}{3}\Rightarrow -1=-\frac{p}{3}\Rightarrow p=3$