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

4 Answers
Mar 25, 2017

Answer:

#m=+-2#

Explanation:

Given:

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

Here's one method:

Suppose the roots are #alpha#, #alpha# and #beta#.

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#

Mar 25, 2017

Answer:

#m in {-2,+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: #y=x^3-3x#
enter image source here
we can see that the relation has turning points at #(-1,2)# and #(1,-2)#
That is the relation has a double root when
#color(white)("XXX")2=x^3-3xcolor(white)("XX")rarrcolor(white)("XX")x^3-3xcolor(magenta)(-2)=0color(white)("XX")rarrcolor(white)("XX")m=color(magenta)(-2)#
and when
#color(white)("XXX")-2=x^3-3xcolor(white)("XX")rarrcolor(white)("XX")x^3-3xcolor(magenta)(+2)=0color(white)("XX")rarrcolor(white)("XX")m=color(magenta)(+2)#

Using Calculus
A slope of zero implies that the derivative is equal to zero at that point.
If #f(x)=x^3-3x+m# for some constant #m#
then the slope is zero when
#color(white)("XXX")(df)/(dx)=3x^2-3=0#
#color(white)("XXXXXXX")x^2=1color(white)("XX")rarrcolor(white)("XX")x=+-1#

Substituting #x=1# and #x=-1# back into the equation:
#color(white)("XXX")x^3-3x+m=0#
gives the solution values for #m#
#color(white)("XXX")m=2color(white)("X")andcolor(white)("X")m=-2#

Mar 25, 2017

Answer:

#m = +-2#

Explanation:

Here's another method using the discriminant...

The discriminant #Delta# of a cubic #ax^3+bx^2+cx+d# is given by the formula:

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

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

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

is in the form:

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

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

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 #Delta = 0# if and only if #m = +-2#

Mar 26, 2017

Answer:

#m=+-2#

Explanation:

Another method

if a cubic has a repeated root #" "alpha" "#then we have:

#P(x)=(x-alpha)^2(x-beta)#

differentiating

#=>P'(x)=2(x-alpha)(x-beta)+(x-alpha)^2#

#:.P'(alpha)=2(alpha-alpha)(alpha-beta)+(alpha-alpha)^2=0#

ie for a repeated root #" "alpha" "#in a cubic #" "P'(alpha)=0#

so we have

#P(X)=X^3-3X-m#

#P'(X)=3X^2-3#

solving#" "3X^2-3=0=>X=+-1#

#X=1=>1^3-3xx1-m=0#

#=>m=-2#

#X=-1=>(-1)^3-3xx(-1)-m=0#

#=>m=2#