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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mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

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

 

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.`

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