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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