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

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

• $p=-6$ does not satisfy $p \gt 0$. Mind that $x_1^3=1$ has more roots than just $x_1=1$. – dxiv Jan 26 '17 at 7:37
• If p > 0 then why do I get p = -6 when I solve in above manner. What is incorrect in the way I arrived at the value of p? – Sunil Jan 26 '17 at 7:44
• This is what is incorrect: $x_{1}^{3} = 1 \implies x_{1} = 1\,$. There are two other roots other than $x_1=1$ which you are ignoring. Either of those two roots actually give $p=3 \gt 0$. – dxiv Jan 26 '17 at 8:07
• @Rumplestillskin typographical error in your factorization? Second factor should be $x^2+x+1$. Can't edit, need to delete and repost. – Oscar Lanzi Jan 26 '17 at 11:23
• Brilliant pun! With regards to the remaining two roots. $x^3−1$ is the difference of two perfect cubes which can be written equivalently as $(x−1)(x^2+x+1)$. @OscarLanzi should be good now. – Rumplestillskin Jan 26 '17 at 15:48

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 $x_1^2+x_1+1=0 \iff x_2+x_1+1=0 \iff x_1+x_2=-1 \iff p=3\,$.

• That is a beautiful answer especially how you went about solving the polynomial equation $x^{3} = 1$ by factoring $x^{3} - 1$. Just too good. – Sunil Jan 26 '17 at 8:38

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!

• Once you get the factorization for $p$ you see that the complicated factor with the square root is strictly positive when $p$ is. So we can ignore it forcing $p=3$. – Oscar Lanzi Jan 26 '17 at 11:30

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 $p=3$.

• How did you come up with the equation $x^{2} + x + \frac{p}{3} = 0$? – Sunil Jan 26 '17 at 8:13
• @Sunil from the last equation, both $\alpha$ and $\beta$ satisfy that quadratic equation.ㅣ – S.C.B. Jan 26 '17 at 8:47
• Ok. I get it now. Thanks. – Sunil Jan 26 '17 at 9:04