# for what range of values of k does the equation 12x+3x^2-2x^3=k have a) Exactly three real roots? b) Two real roots? c) Only one real root?

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### 2 Answers

We need to find where the turning points are of the cubic on the lefthand side of the equation. We do this by differentiating the cubic with respect to `x`:

`d/(dx) (12x + 3x^2 - 2x^3) = 12 + 6x - 6x^2`

To find the turning points we solve

`-6x^2 + 6x + 12 = 0`

ie, `x^2 - x - 2 = 0`

By insepction

`(x+1)(x-2) = 0`

So the turning points are at `x = -1` and `x = 2`

When `k` is between these two turning points, the equation has three real roots, when `k` is equal to either of these turning points, there are two roots (one is repeated)` `. Otherwise, `k` has only 1 real root.

**k has one real root when k < -1 or k >2**

**k has two real roots when k= -1 or k=2**

**k has three real roots when -1 < k < 2**

You need to consider k=0 such that:

`12x+3x^2-2x^3=0`

Factoring out x yields:

`x(12 + 3x - 2x^2) = 0`

You need to solve the following equations such that:

`x = 0`

`12 + 3x - 2x^2 = 0 => 2x^2 - 3x - 12 = 0`

Using quadratic formula yields:

`x_(1,2) = (-(-3)+-sqrt((-3)^2 - 4*2*(-12)))/(2*2)`

`x_(1,2) = (3+-sqrt(9+96))/4 => x_(1,2) = (3+-sqrt(105))/4`

**Hence, considering k = 0, the equation will have three real roots,`x = 0` , `x = (3-sqrt(105))/4` and `x = (3+sqrt(105))/4.` **