<s>[INST] Solve a cubic polynomial? I've been having trouble with this question:
Solve the equation, $$5x^3 - 24x^2 + 9x + 54 = 0$$  given that two of its roots are equal.
I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but I am still unsuccessful.
Any help would be greatly appreciated.
 [/INST] By Vieta's formula, $$\begin{cases}2a+b=\dfrac{24}5\\a^2+2ab=a(a+2b)=\dfrac95\\a^2b=-\dfrac{54}5 \end{cases}$$
Notice that $0$ is not a root of the original equation, so both $a,b$ are nonzero. Divide the second equation by the third equation to get $\dfrac{a+2b}{ab}=\dfrac2a
+\dfrac1b=-\dfrac1{6}$. Multiplying by the first equation, we have $$(2a+b)(\dfrac2a
+\dfrac1b)=5+2(\dfrac ba+\dfrac ab)=-\dfrac45 $$
Let $\dfrac ba=t$, then the above equation becomes $$5+2(t+\dfrac1t)=-\dfrac45\implies 10t^2+29t+10=(5t+2)(2t+5)=0$$
Hence $t=-\dfrac25$ or $-\dfrac52$. If $t=-\dfrac25, 2a+b=2a+at=\dfrac85a=\dfrac{24}5,$ so $a=3$. On the other hand, if $t=-\dfrac52, 2a+b=-\dfrac12a=\dfrac{24}5,$ so $a=-\dfrac{48}5$. But from the third equation we must have $b\lt 0$, and then $a,b$ are both $\lt 0$, so $2a+b\lt 0$ and then the first equation cannot hold. Therefore $a=3, b=-\dfrac{6}5$ are the roots of the original cubic.
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