Question

How do you solve `x( x - 5) > 6`?

Answer 1

`x>6`, `x< -1`

Explanation:

Let's distribute the `x` on the left side to get

`x^2-5x>6`

We have a second-degree term, so this is beginning to look like a quadratic. To put it in a more traditional form, let's subtract `6` from both sides.

`x^2-5x-6>0`

Let's see if we can factor the left. What two numbers sum up to `-5` and have a product of `-6`?

After some trial and error, we arrive at `-6` and `1`. This allows us to factor this as

`(x-6)(x+1)>0`

We have two scenarios:

`x-6>0` and `x+1>0`

We can solve them simultaneously to get

`x>6` and `x> -1`

We essentially have two conflicting scenarios. Let's think for a minute. We had the following:

`(x-6)(x+1)>0`

If the product of two things is greater than zero (positive), they must have the same signs.

`x-6` is negative if `x<6`, and positive if `x>6`.

`x+1` is negative if `x<-1`, and positive if `x> -1`

When is `(x-6)(x+1)` positive?

If `x=-1` or `6`, we will be equal to zero, but we are positive when `x>6` and if `x< -1`. These are our solutions.

If the latter part of my solution seemed confusing, what I did is essentially a sign chart. More information can be found [here.](https://socratic.org/questions/what-is-a-sign-chart)

Hope this helps!