Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$?

Question

If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must

be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$

My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula

$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$

$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$

$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$

Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.

Help is required.

Thanks

Answer 1

Since $\beta=-2\alpha-\frac{3b}{a}$, we have

$${\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\Rightarrow a{\alpha}^2+6b\alpha+c=0\Rightarrow 2a{\alpha}^3+12b{\alpha}^2+2c\alpha=0,$$ $${\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.$$

Hence, we have $$9{\alpha}^2+2c\alpha+d=0\Rightarrow 9a{\alpha}^2+2ca\alpha+ad=0.$$ Hence, we've already have $$9a{\alpha}^2+54b\alpha+9c=0,$$ we have $$(2ca-54b)\alpha=9c-ad$$ Hence, we have $$\alpha=\frac{9c-ad}{2(ca-27b)}.$$

I don't know how to reach your value.

Answer 2

From 1 , take $\beta = -\frac{3b}a - 2\alpha$

Substitute $\beta$ in 3,

$$\alpha^2 \left(-\frac{3b}a - 2\alpha\right) = -\frac da$$

You will get a cubic in $\alpha$ which you can solve.

Or you can substitute the value of $\beta$ in 2 from where you will get a quadratic in $\alpha$ which will be easier to solve.

Answer 3

First Answer: \begin{align*} 2\alpha +\beta = -\frac{3b}{a} &\Rightarrow \beta=-\frac{3b}{a}-2\alpha\\ &\Rightarrow {\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\\ &\Rightarrow -3\alpha^2-\frac{6b}{a}\alpha=\frac{3c}{a} \\ &\Rightarrow a{\alpha}^2+2b\alpha+c=0\tag{0}\\ &\Rightarrow 2a{\alpha}^3+4b{\alpha}^2+2c\alpha=0\tag{1} \end{align*} and \begin{align*} {\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.\tag{2} \end{align*}

By subtracting (1) and (2) we have $$b\alpha^2+2c\alpha+d=0\Rightarrow ab\alpha^2+2ac\alpha+ad=0.$$ Now from (0) we have $$ab\alpha^2+2b^2\alpha+bc=0$$ Now subtract two last equations: $(2ac-2b^2)\alpha+ad-bc=0$ Thus $$\fbox{$\alpha=\frac{bc-ad}{2(ac-b^2)}$}$$

Second Answer for alpha:

You can continue like this $$ 2\alpha +\beta = -\frac{3b}{a} \Rightarrow \beta=-\frac{3b}{a}-2\alpha\\ $$ put it in the last one: \begin{align*} \left(-\frac{3b}{a}-2\alpha\right)\alpha^2=-\frac{d}{a}&\Rightarrow-2\alpha^3-\frac{3b}{a}\alpha^2=-\frac{d}{a}\\ &\Rightarrow 2a\alpha^3+3b\alpha^2-d=0\\ \end{align*} on the other hand $\alpha$ must satisfy the original equation, thus $$a\alpha^3+3b\alpha^2+3c\alpha+d=0$$ Then add two last equations: $$3a\alpha^3+6b\alpha^2+3c\alpha=0$$ Since $\alpha\ne0$ we have : $a\alpha^2+2b\alpha+c=0$, thus $$\alpha=-b+\sqrt{b^2-ac}\quad \text{ or }\quad \alpha=-b-\sqrt{b^2-ac} $$