Roots of the equation $x^3+15x^2+cx+860=0$

Question

If $-5+i\beta$ , $-5+i\gamma$ ,$\beta^2\ne\gamma^2$; $\beta,\gamma \in R$ are the roots of the equation $x^3+15x^2+cx+860=0$, $c\in R$, then find the three roots of the equation.

My approach is as follow, It is mentioned that the roots are not conjugate. Hence the third roots is $-5-i(\beta + \gamma)$. The product is $(-5+i\beta)(-5+i\gamma) (-5-i(\beta + \gamma)=-860$,

I am not able to proceed from here.

Answer 1

From Vieta's formulas, we know that the three roots $r$,$s$ and $t$ satisfy the following equations:

• $r+s+t=-15$
• $rs+st+tr=c$
• $rst=-860$

Assume now, according to the information we have, $r=-5+i\beta$ and $s=-5+i\gamma$. From the first formula, we obtain:

$$ t = -5 - i(\beta+\gamma) $$

Since a cubic equation has at least one real root, we must choose or $\beta$ or $\gamma$ zero to satisfy the condition $\beta^2\ne\gamma^2$. So assume $\beta=0$.

The last equation gives us now:

$$ rst = -5\cdot(-5+i\gamma)(-5-i\gamma) = -5\cdot(25+\gamma^2)=-860$$

Which gives us $\gamma=7\sqrt3$.

So, the roots are $r=-5$, $s=-5+7\sqrt3$ and $t=-5-7\sqrt3$. The value of $c$ can now be obtained from the second equation and gives $c=222$.

Answer 2

Since $\beta^2\not=\gamma^2$, we get $\beta+\gamma\not=0$.

It follows from what you've done that the third root is not real.

Using the fact that there is at least one real root, we see that either $\beta=0$ or $\gamma=0$.

Since $x=-5$ is a root, we have $$(-5)^3+15\cdot (-5)^2+c\cdot (-5)+860=0$$ from which $c=222$ follows.

Dividing $x^3+15x^2+222x+860$ by $x+5$ gives $$x^2+10x+72$$

So, the roots are $x=-5,-5\pm 7\sqrt 3\ i$.