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

Dec 9, 2015

$z < - 3$ or $z > 4$

#### Explanation:

First, bring everything to the left side:

$z \left(2 - z\right) < z - 12$

... subtract $z$ from both sides...

$\iff z \left(2 - z\right) - z < - 12$

... subtract $12$ from both sides ...

$\iff z \left(2 - z\right) - z + 12 < 0$

Expand the expression on the left side and simplify:

$\iff - {z}^{2} + z + 12 < 0$

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!

$\iff {z}^{2} - z - 12 > 0$

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

$z = \frac{1 \pm \sqrt{1 + 4 \cdot 12}}{2} = \frac{1 \pm 7}{2}$

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).