Find the values of '#k#' if equation #x^3-3x^2+2=k# has:- (i)3 real roots (ii)1 real root?

1 Answer
George C.
Aug 11, 2018

(i) The given equation has #3# real roots when #k in (-2, 2)#

(ii) The given equation has #1# real root when #k in (-oo, -2) uu (2, oo)#

Explanation:

We could solve this with the aid of the cubic discriminant, but let's look at it without...

Given:

#x^3-3x^2+2=k#

Let:

#f(x) = x^3-3x^2+2-k#

First note that if #k=0# then #x=1# is a real zero and #(x-1)# a factor:

#x^3-3x^2+2 = (x-1)(x^2-2x-2)#

#color(white)(x^3-3x^2+2) = (x-1)(x^2-2x+1-3)#

#color(white)(x^3-3x^2+2) = (x-1)((x-1)^2-(sqrt(3))^2)#

#color(white)(x^3-3x^2+2) = (x-1)(x-1-sqrt(3))(x-1+sqrt(3))#

which has #3# real zeros.

Next note that #f(x)# will have a repeated zero if and only if it has a common factor with:

#f'(x) = 3x^2-6x = 3x(x-2)#

If #x=0# is a root then #k=2#

If #x=2# is a root then #k=-2#

These two values of #k# split the real line into #3# parts:

#(-oo, -2)" "(-2, 2)" "(2, oo)#

Since we have observed that #f(x)# has #3# zeros when #k=0#, we can deduce that:

(i) The given equation has #3# real roots when #k in (-2, 2)#

(ii) The given equation has #1# real root when #k in (-oo, -2) uu (2, oo)#

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