How to solve this? Find m in RR  for which X^3-3X+m=0 has a double root.

Mar 25, 2017

$m = \pm 2$

Explanation:

Given:

${x}^{3} - 3 x + m = 0$

Here's one method:

Suppose the roots are $\alpha$, $\alpha$ and $\beta$.

Then:

${x}^{3} - 3 x + m = \left(x - \alpha\right) \left(x - \alpha\right) \left(x - \beta\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 x + m} = {x}^{3} - \left(2 \alpha + \beta\right) {x}^{2} + \left({\alpha}^{2} + 2 \alpha \beta\right) x - {\alpha}^{2} \beta$

Equating coefficients, we have:

$\left\{\begin{matrix}2 \alpha + \beta = 0 \\ {\alpha}^{2} + 2 \alpha \beta = - 3 \\ - {\alpha}^{2} \beta = m\end{matrix}\right.$

From the first equation, we have:

$\beta = - 2 \alpha$

Substituting this into the second and third equations, we find:

$\left\{\begin{matrix}- 3 {\alpha}^{2} = - 3 \\ 2 {\alpha}^{3} = m\end{matrix}\right.$

From the first of these equations we find:

${\alpha}^{2} = 1$

and hence:

$\alpha = \pm 1$

Then from the second:

$m = 2 {\alpha}^{3} = \pm 2$

Mar 25, 2017

$m \in \left\{- 2 , + 2\right\}$

Explanation:

There are a couple ways to approach this once we recognize that "having a double root" is equivalent to having a slope of zero.

Graphically, if we consider the simpler relation: $y = {x}^{3} - 3 x$

we can see that the relation has turning points at $\left(- 1 , 2\right)$ and $\left(1 , - 2\right)$
That is the relation has a double root when
$\textcolor{w h i t e}{\text{XXX")2=x^3-3xcolor(white)("XX")rarrcolor(white)("XX")x^3-3xcolor(magenta)(-2)=0color(white)("XX")rarrcolor(white)("XX}} m = \textcolor{m a \ge n t a}{- 2}$
and when
$\textcolor{w h i t e}{\text{XXX")-2=x^3-3xcolor(white)("XX")rarrcolor(white)("XX")x^3-3xcolor(magenta)(+2)=0color(white)("XX")rarrcolor(white)("XX}} m = \textcolor{m a \ge n t a}{+ 2}$

Using Calculus
A slope of zero implies that the derivative is equal to zero at that point.
If $f \left(x\right) = {x}^{3} - 3 x + m$ for some constant $m$
then the slope is zero when
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{df}}{\mathrm{dx}} = 3 {x}^{2} - 3 = 0$
$\textcolor{w h i t e}{\text{XXXXXXX")x^2=1color(white)("XX")rarrcolor(white)("XX}} x = \pm 1$

Substituting $x = 1$ and $x = - 1$ back into the equation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} - 3 x + m = 0$
gives the solution values for $m$
$\textcolor{w h i t e}{\text{XXX")m=2color(white)("X")andcolor(white)("X}} m = - 2$

Mar 25, 2017

$m = \pm 2$

Explanation:

Here's another method using the discriminant...

The discriminant $\Delta$ of a cubic $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

The cubic will have repeated roots if and only if $\Delta = 0$.

${x}^{3} - 3 x + m = 0$

is in the form:

$a {x}^{3} + b {x}^{2} + c x + d = 0$

with $a = 1$, $b = 0$, $c = - 3$ and $d = m$

So:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

$\textcolor{w h i t e}{\Delta} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(0\right)}^{2} {\left(- 3\right)}^{2}}}} - 4 \left(1\right) {\left(- 3\right)}^{3} - \textcolor{red}{\cancel{\textcolor{b l a c k}{4 {\left(0\right)}^{3} \left(m\right)}}} - 27 {\left(1\right)}^{2} {\left(m\right)}^{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{18 \left(1\right) \left(0\right) \left(- 3\right) \left(m\right)}}}$

$\textcolor{w h i t e}{\Delta} = 4 \cdot 27 - 27 {m}^{2}$

$\textcolor{w h i t e}{\Delta} = 27 \left({2}^{2} - {m}^{2}\right)$

$\textcolor{w h i t e}{\Delta} = 27 \left(2 - m\right) \left(2 + m\right)$

So $\Delta = 0$ if and only if $m = \pm 2$

Mar 26, 2017

$m = \pm 2$

Explanation:

Another method

if a cubic has a repeated root $\text{ "alpha" }$then we have:

$P \left(x\right) = {\left(x - \alpha\right)}^{2} \left(x - \beta\right)$

differentiating

$\implies P ' \left(x\right) = 2 \left(x - \alpha\right) \left(x - \beta\right) + {\left(x - \alpha\right)}^{2}$

$\therefore P ' \left(\alpha\right) = 2 \left(\alpha - \alpha\right) \left(\alpha - \beta\right) + {\left(\alpha - \alpha\right)}^{2} = 0$

ie for a repeated root $\text{ "alpha" }$in a cubic $\text{ } P ' \left(\alpha\right) = 0$

so we have

$P \left(X\right) = {X}^{3} - 3 X - m$

$P ' \left(X\right) = 3 {X}^{2} - 3$

solving$\text{ } 3 {X}^{2} - 3 = 0 \implies X = \pm 1$

$X = 1 \implies {1}^{3} - 3 \times 1 - m = 0$

$\implies m = - 2$

$X = - 1 \implies {\left(- 1\right)}^{3} - 3 \times \left(- 1\right) - m = 0$

$\implies m = 2$