Is answer incorrect for this quadratic equation?

Question

I am solving the following question on quadratic equation and I always get $p = -6$ as answer but the answer in book is $p = 3$.

Question

What am I missing in my solution or is the answer mentioned in book incorrect?

Problem

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

My Solution

Let $x_{1}$ and $x_{2}$ be the roots of this equation.

So, $x_{1} + x_{2} = -\frac{p}{3}$

Also, $x_{1}. x_{2} = 1 =>x_{1}^{3} = 1 => x_{1} = 1$

Therefore, substituting in first equation we get

$1 + 1^{2} = -\frac{p}{3} => p = -6$

Answer 1

Let the roots be $x_{1,2}$ and assume $x_2=x_1^2$. Then by Vieta's formulas $x_1x_2=1=x_1^3$.

The equation $x_1^3=1 \iff x_1^3-1=0 \iff (x_1 -1)(x_1^2+x_1+1) = 0$ has $3$ roots:

• $x_1=1\,$, but substituting this back into the equation gives $3 + p + 3 = 0$ i.e. $p=-6 \lt 0\,$ which doesn't satisfy the requirement $p \gt 0$.

• $x_1 = (-1 \pm i \sqrt{3}) / 2\,$, otherwise known as the complex cube roots of unity, which satisfy $x_1^2+x_1+1=0 \iff x_2+x_1+1=0 \iff x_1+x_2=-1 \iff p=3\,$.

Answer 2

Let $\alpha$, $\beta$ be roots of $3x^{2}+ px + 3 = 0$.

Note they satisfy $x^2+\frac{p}{3}x +1$. From Vieta's Formula, we have $$\alpha+\beta=-\frac{p}{3}<0$$ $$\alpha \beta=1$$

Now, if $\alpha^2=\alpha$ and $\beta^2=\beta$,then we have $\{\alpha, \beta\} \subseteq \{0,1\}$. A contradiction, as the sum of $\alpha$ and $\beta $ is negative.

So, we have that $$\alpha^2=\beta=-\frac{p}{3}-\alpha \implies \alpha^2+\alpha+\frac{p}{3}=0 $$$$\beta^2=\alpha=-\frac{p}{3}-\beta \implies \beta^2+\beta+ \frac{p}{3}=0$$ So $\alpha$, $\beta$ are solutions to $x^2+x+\frac{p}{3}$.

By Vieta's formula, $\alpha \beta=\frac{p}{3}=1$ So $p=3$.

Answer 3

Another method - although not very pretty is to do the following, the two roots $x_1$ and $x_2$ are given by $$ x_{1,2} = \frac{1}{6} \bigg( -p \pm \sqrt{p^2-36} \bigg).$$

We also know that $x_1=x_2^2$. This yields $$ \frac{1}{18}p^2-\frac{1}{18}p\sqrt{p^2-36}-1 = -\frac{1}{6}p-(1/6)\sqrt{p^2-36} $$

which can be simplified to give

$$ -\frac{1}{18}(p-3)\bigg(p - \sqrt{p^2-36}+6\bigg)=0.$$

Either $p=3$ or $p=-6$. We know that $p>0$ so the second root can be ignored!