Question

For the equation $3x^2+px+3=0,p>0$, if one of the root is square of the other, then $p$ is equal to

• A. $1/3$
• B. $1$
• C. $3$
• D. $2/3$

Answer 1

Answer: (C)

$3$

$3x^2+px+3=0$

Here, $a=3,b=p,c=3$

Let $m$ be the one root

Another root $=m^2$

Therefore,

Sum of roots $=\frac{-b}{a}$

$\therefore m^2+m=\frac{-p}{3}.....\left(1\right)$

Product of roots $=\frac{c}{a}$

$m^2\cdot m=\frac{3}{3}$

$m^3=1$

$\Rightarrow m=1,\omega ,{\omega }^2$

Case I:- For $m=\omega$ or ${\omega }^2$

$\omega +{\omega }^2=\frac{-p}{3}$

$-1=\frac{-p}{3}$

$\Rightarrow p=3$

Case II:- For $m=1$

$1+1=\frac{-p}{3}$

$\Rightarrow p=-6$

$\because P>0\left(\text{Given}\right)$

$\therefore p=3$