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

##### 4 Answers

#### Explanation:

Given:

#x^3-3x+m = 0#

Here's one method:

Suppose the roots are

Then:

#x^3-3x+m = (x-alpha)(x-alpha)(x-beta)#

#color(white)(x^3-3x+m) = x^3-(2alpha+beta)x^2+(alpha^2+2alphabeta)x-alpha^2beta#

Equating coefficients, we have:

#{ (2alpha+beta = 0), (alpha^2+2alphabeta = -3), (-alpha^2beta = m) :}#

From the first equation, we have:

#beta = -2alpha#

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

#{ (-3alpha^2 = -3), (2alpha^3 = m) :}#

From the first of these equations we find:

#alpha^2 = 1#

and hence:

#alpha = +-1#

Then from the second:

#m = 2alpha^3 = +-2#

#### 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:

we can see that the relation has turning points at

That is the relation has a double root when

and when

**Using Calculus**

A slope of zero implies that the derivative is equal to zero at that point.

If

then the slope is zero when

Substituting

gives the solution values for

#### Explanation:

Here's another method using the discriminant...

The discriminant

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

The cubic will have repeated roots if and only if

#x^3-3x+m = 0#

is in the form:

#ax^3+bx^2+cx+d = 0#

with

So:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

#color(white)(Delta) = color(red)(cancel(color(black)((0)^2(-3)^2)))-4(1)(-3)^3-color(red)(cancel(color(black)(4(0)^3(m))))-27(1)^2(m)^2+color(red)(cancel(color(black)(18(1)(0)(-3)(m))))#

#color(white)(Delta) = 4*27 - 27m^2#

#color(white)(Delta) = 27(2^2-m^2)#

#color(white)(Delta) = 27(2-m)(2+m)#

So

#### Explanation:

Another method

if a cubic has a repeated root

differentiating

ie for a repeated root

so we have

solving