Question

For the equation $3x^2+px+3=0,p<0$ if one of the roots is square of the other, then $p$ is eqal to:

• A. $\frac{1}{3}$
• B. $1$
• C. $-6$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $-6$

Given equation is

$3x^2+px+3=0,p>0$

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

If one of the roots is square of the other, then

$(a{c}^2{)}^{\frac{1}{3}}+(a^{2}c{)}^{\frac{1}{3}}+b=0$

$⟹(27{)}^{\frac{1}{3}}+(27{)}^{\frac{1}{3}}+p=0$

$⟹3+3+p=0$

$\therefore p=-6$