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.
$$(x-a)^2(x-b) = (x^2-2ax+a^2)(x-b)=x^3-(2a+b)x^2+a(a+2b)x-a^2b$$ $$=x^3-\frac{24}{5}x^2+\frac{9}{5}x+\frac{54}{5}$$
Then $$2a+b=\frac{24}{5}$$ $$a+2b=\frac{9}{5a}$$
Multiply the first by 2 and subtract: $$3a=\frac{48}{5}-\frac{9}{5a}$$
Now it should not be difficult to find $a$. Of course this equation gives two possible values of $a$, and only one is right (you have to check with your degree three equation).
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.
