Question

For the equation $3x^2+px+3=0$, the roots are real and $1$ root is the square of the other, Find $p$?

Answer 1

Step 1: Form the equation

The given equation is $3x^2+px+3=0$

where $a=3,b=p$ and $c=3$

Let us assume the roots of the given equation are $y$ and $y^2$.

As per the given equation

Sum of the roots $-\frac{b}{a}$ is

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

Product of the roots $\frac{c}{a}$ is

$y^3=1$

$\Rightarrow y=1$

Step 2: Calculate $p$

Substituting the value of $y$ on equation $\left(1\right)$,

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

$\Rightarrow$$p=-6$

Hence the value of $p$ is $-6$.