Question

How do you solve `z(2 - z) < z - 12`?

Answer 1

`z< -3` or `z>4`

Explanation:

First, bring everything to the left side:

`z(2-z) < z - 12`

... subtract `z` from both sides...

`<=> z(2-z) - z < - 12`

... subtract `12` from both sides ...

`<=> z(2-z) - z + 12 < 0`

Expand the expression on the left side and simplify:

`<=> - z^2 + z + 12 < 0`

At this point, I would recommend to multiply by `-1` on both sides in order to gain the factor `1` in front of the `z^2` term.

Be careful though: if you multiply by a negative factor or divide by a negative factor, you need to flip the inequality sign!

`<=> z^2 - z - 12 > 0`

Now, let's find the roots of the quadratic expression at the left side. It can be done e.g. with the quadratic formula:

`z = (1 +- sqrt(1 + 4 * 12))/(2) = (1 +-7)/2`

So, the roots are `z = 4` and `z = -3`.

Now, let's think about what the inequality `z^2 - z - 12 > 0` means.

• It's a quadratic function that opens up since the coefficient of `z^2` is positive.
• Further, its roots are `z = 4` and `z = -3` which means that the value of the function is `0` for those `z` values.
• This means that the function has positive values if `z> 4` or `z< -3` hold.

It can also help seeing this if you look at the graph:

graph{x^2 - x - 12 [-10, 10, -15, 15]}

As we have a `> 0` inequality, the part we are really interested in is the one above the `x`-axis:

So, the solution is

`z < -3` or `z > 4`

or, depending on your notation preference,

`z in (- oo; -3) uu (4; oo)`.