Question

Find the set of values of k,where k is a constant,when equation x^3-12x^2+45x-34=k has (a) one root (b) three roots?

Answer 1

`color(red)((a))`, `20 < k < 16`

`color(red)((b))`, `16 < k < 20`

Explanation:

.

`y=x^3-12x^2+45x-34=k`

We know that this is a cubic function and ordinarily would have three roots, i.e. it would cross the `x`-axis in three points.

We also know that a cubic function typically has one maximum and one minimum unless the function is a perfect cubic in which case it would cross the `x`-axis in one point which would be its inflection point without any maximum or minimum.

A perfect cubic function would be in the form of:

`y=(x-a)^3`

If we expand this we would get:

`y=x^3-3ax^2+3a^2x-a^3`

Setting the corresponding coefficients equal to each other, we get:

`-3a=-12, 3a^2=45, and -a^3=-34-k`

This set of three equation does not have a solution as is evident. Therefore, our function can not be a perfect cubic. This means it has a minimum and a maximum and an inflection point.

`color(red)((a))`

In order for it to have one root, it would have to cross the `x`-axis in one point only. This means that its minimum and maximum points would have to both be either above the `x`-axis or below the `x`-axis.

Let's find these points. We will take the derivative of the function and set it equal to `0`:

`dy/dx=3x^2-24x+45=3(x^2-8x+15)`

`x^2-8x+15=0`

`(x-3)(x-5)=0`

`x=3 and 5`

Let's plug them into the function to find their `y` values:

`x=3, :. y=20-k`

`x=5, :. y=16-k`

For these points to be above the `x`-axis, their `y` values have to be positive:

`y=20-k>0` and `y=16-k>0, :. K<20 and k<16`

We have to go with `k<16` to keep both points above the `x`-axis. The following graph is drawn for `k=15`:

For the points to be below the `x`-axis, their `y` values would have to be negative:

`y=20-k<0 and y=16-k<0, :. k>20 and k>16`

We have to go with `k>20` to keep both points below the `x`-axis. The following graph is drawn for `k=21`:

Therefore, the condition for one root is:

`20 < k < 16`

`color(red)((b))`

For the function to have three roots, `k` would have to be outside the range specified in `color(red)((a))`.

Please note that if `k=16 or 20` the function will have two roots because the graph will have either the maximum or the minimum touch the `x`-axis and not cross it

The following graph is drawn for `k=16`:

The following graph is drawn for `k=20`:

Therefore,

`16 < k < 20` for the function to have three roots. The following graph is drawn for `k=18`: