Question

How do you solve `\frac { 1} { 2} x ^ { 2} - x > 4`?

Answer 1

`x in (–oo, –2)uu(4,oo)`.
or
`SS={x in RR|x<–2 or x>4}`.

Explanation:

Bring everything to one side.

`1/2 x^2 - x - 4 > 0`

Factor that side if possible.
Factor the `1/2` out of all three terms.

`1/2 (x^2 - 2x - 8) > 0`

Recognize that
`–4 xx 2 = 8,` and
`–4 + 2 = –2.`

`1/2(x-4)(x+2)>0`

Divide both sides by `1/2`:

`(x-4)(x+2)>0`

We now have a product of two factors on the left: `x-4` and `x+2`. We are interested in when this product is positive `(>0).`

Both factors depend on `x`. Therefore, their product will be positive when `x` is small/large enough to make the factors either both negative or both positive.

The 1st factor, `x-4`, is negative when `x<4`.
The 2nd factor, `x+2`, is negative when `x<–2`.

So both factors will be negative when `x<–2`.

Similarly, both factors will be positive when `x>4`.

Our solution is all `x` in both these regions: `x<–2 uu x>4.`