Question

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

Answer 1

(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)`