Animated Solution for Mathematics - Quadratic Equations: 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

We are given the quadratic equation $3x^2+px+3=0$. We are tasked with finding the value of $p$ under the constraint that $p>0$, given that one root is the square of the other.

Let the roots of the equation be $\alpha$ and ${\alpha }^2$.

We utilize Vieta's formulas for a quadratic equation $a{x}^2+bx+c=0$, where the product of the roots is $\frac{c}{a}$ and the sum is $-\frac{b}{a}$.

For our specific equation $3x^2+px+3=0$, the product of the roots is:

This yields the elegant condition ${\alpha }^3=1$.

The equation ${\alpha }^3=1$ provides three possible values for $\alpha$: the real root $1$ and the complex roots $\omega$ and ${\omega }^2$, where $\omega =e^{i\frac{2\pi }{3}}$.

If $\alpha =1$, then the roots are $1$ and $1^2=1$. The sum of the roots is $1+1=2$.

According to Vieta's formulas, the sum of the roots is also $-\frac{p}{3}$. Setting these equal:

However, the problem explicitly states the constraint $p>0$. Since $-6$ is not greater than $0$, we must reject this solution.

If $\alpha =\omega$, then the roots are $\omega$ and ${\omega }^2$. We utilize the fundamental property of the cube roots of unity:

Returning to the sum of the roots equation, we have:

Substituting the complex sum $\omega +{\omega }^2=-1$ into the equation:

Solving for $p$, we find:

We verify our result against the initial constraint $p>0$. Since $3>0$, the condition is satisfied.

The value of $p$ is 3.