If $\alpha$ be a multiple root of the order 3 of the equation $x⁴+bx²+cx+d=0$ then $α= 8d/3c​$

Question

I found this statement online here and I think this statement is false.I would first prove a general statement and then I would use it to find $\alpha$ in order to discard the statement given in online link.

Theorem:If the equation $x^4+ax^3+bx^2+cx+d=0$ has three equal roots, show that each of them is equal to $\frac{6c-ab}{3a^2-8b}$​

Proof;Three roots of the given equation are same hence, the roots can be assumed as $\alpha,\alpha,\alpha,\beta$

Here, $S_1=3α+β=−a;S_2 =3α(α+β)=b;S_3=α^2(α+3β)=−c;S_4=α^3β=d$

We need to evaluate the value of $\frac{6c-ab}{3a^2-8b}$​.

$6c−ab=α(3α^2−6αβ+3β^2)$

$3a^2−8b=3α^2−6αβ+3β^2$

Therefore,$\frac{6c-ab}{3a^2-8b}=\frac{α(3α^2−6αβ+3β^2)}{(3α^2−6αβ+3β^2)}=\alpha$

Now according to this ,putting $a=0$ we get $\alpha=\frac{-3c}{4b}$ which is not same as $\frac{8d}{3c}$ as claimed in online site.

My question is: Is the claim in online site wrong?or I have gone wrong somewhere in my proof.

Any help is much appreciated!

Answer 1

Let the roots be $\alpha, \alpha, \alpha, - 3\alpha$ so that their sum is $0$.

Now sum of products of 3 roots taken at a time is $$\alpha^3-3\alpha^3-3\alpha^3-3\alpha^3$$ and this must equal $-c$ and hence $\alpha^3=c/8$.

Now product of roots is $-3\alpha^4=d$ so that $\alpha^4=-d/3$ and hence $\alpha=(-d/3)/(c/8)=-8d/3c$.

Your question has thus a sign error. You can convince yourself by putting $\alpha=1$. The desired polynomial is $$(x-1)^3(x+3)=x^4-6x^2+8x-3$$ so that $c=8,d=-3$ and $-8d/3c=1=\alpha$.

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I checked your approach and your theorem in question and its proof are correct. Thus $\alpha $ should equal $-3c/4b$.

Let's observe that sum of product of roots taken $2$ at a time is $$\alpha^2+\alpha^2+\alpha ^2-3\alpha ^2-3\alpha ^2-3\alpha^2$$ and this should equal $b$. Thus $b=-6\alpha^2$ and hence $$-\frac{3c}{4b}=-\frac{24\alpha^3}{(-24\alpha^2)}=\alpha$$ So we have $$\alpha=-\frac{3c}{4b}=-\frac{8d}{3c}$$

Answer 2

Your answer looks correct to me.

We have,

$$(x-m)^3(x-n)=x^4+bx^2+cx+d=0.$$

$$x^4+bx^2+cx+d=x^4-x^3(3m+n)+x^2(3m^2+3mn)-x(m^3+3m^2n)+m^3n$$ $$\begin{align}\begin{cases}3m=-n \\3m(m+n)=b \\m^2(m+3n)=-c \\ m^3n=d \end{cases} &\implies \begin{cases}-6m^2=b\\8m^3=c\end{cases}\\ &\implies \begin{cases}m=-\frac{3c}{4b},~b≠0\\ n=\frac{9c}{4b},~b≠0\end{cases}\end{align}$$

But, remember that there is a relationship between the coefficients. This implies that the coefficients are not independent. Therefore, to find the root we are looking for, we will use the coefficients that work for us. However, both answers are absolutely correct.

$$\begin{align}\begin{cases}3m=-n \\3m(m+n)=b \\m^2(m+3n)=-c \\ m^3n=d \end{cases} &\implies \begin{cases}-3m^4=d\\8m^3=c\end{cases}\\ &\implies\begin{cases} m=-\frac{8d}{3c}\\ n=\frac{8d}{c}\end{cases}\end{align}$$

This means,

$$\begin{align}\frac{3c}{4b}&=\frac{8d}{3c}\\ \implies c^2&=\frac{32}{9}bd.\end{align}$$

Answer 3

Verification of the Theorem

Expanding $$ (x-u)^3(x-v)=x^4\ \overbrace{-(3u+v)}^{\large a}x^3\ \overbrace{+3u(u+v)}^{\large b}x^2\ \overbrace{-u^2(u+3v)}^{\large c}x\overbrace{\ +u^3v\ \ }^{\large d}\tag1 $$ The first two derivatives are $0$ at $u$ $$ u^4+au^3+bu^2+cu+d=0\tag2 $$ $$ 4u^3+3au^2+2bu+c=0\tag3 $$ $$ 12u^2+6au+2b=0\tag4 $$ Substituting $v=d/u^3$ into $c+u^2(u+3v)=0$ gives $$ u^4+cu+3d=0\tag5 $$ Writing $(2)$-$(5)$ as a matrix equation yields $$ \begin{align} 0 &=\begin{bmatrix} 1&-\frac a4&\frac{3a^2-4b}{48}&-1 \end{bmatrix} \begin{bmatrix} 1&a&b&c&d\\ 0&4&3a&2b&c\\ 0&0&12&6a&2b\\ 1&0&0&c&3d \end{bmatrix} \begin{bmatrix} u^4\\u^3\\u^2\\u\\1 \end{bmatrix}\tag{6a} \\ &=\begin{bmatrix} 0&0&0&\frac{3a^3-8ab}8&\frac{3a^2b-4b^2-6ac-48d}{24} \end{bmatrix} \begin{bmatrix} u^4\\u^3\\u^2\\u\\1 \end{bmatrix}\tag{6b} \end{align} $$ Since $b^2+12d=3ac$, $\text{(6b)}$ gives $$ \begin{align} u &=\frac{3a^2b-4b^2-6ac-48d}{24ab-9a^3}\tag{7a}\\ &=\frac{3a^2b-18ac}{24ab-9a^3}\tag{7b}\\ &=\frac{ab-6c}{8b-3a^2}\tag{7c} \end{align} $$

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When $\boldsymbol{a=0}$

Plugging $a=0$ into $\text{(7c)}$ gives $$ u=-\frac{3c}{4b}\tag8 $$ If we use $a=0$, that is, $v=-3u$, we get $b=-6u^2$, $c=8u^3$, and $d=-3u^4$, we get not only that $u=-\frac{3c}{4b}$, as given in $(8)$, but also that $$ u=-\frac{8d}{3c}\tag9 $$ which says that the problem in question is a factor of $-1$ off.