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

1 Answer
Lotusbluete · Ernest Z.
Dec 9, 2015

z< -3z<−3 or z>4z>4

Explanation:

First, bring everything to the left side:

z(2-z) < z - 12z(2−z)<z−12

... subtract zz from both sides...

<=> z(2-z) - z < - 12⇔z(2−z)−z<−12

... subtract 1212 from both sides ...

<=> z(2-z) - z + 12 < 0⇔z(2−z)−z+12<0

Expand the expression on the left side and simplify:

<=> - z^2 + z + 12 < 0⇔−z2+z+12<0

At this point, I would recommend to multiply by -1−1 on both sides in order to gain the factor 11 in front of the z^2z2 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⇔z2−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)/2z=1±√1+4⋅122=1±72

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

Now, let's think about what the inequality z^2 - z - 12 > 0z2−z−12>0 means.

  • It's a quadratic function that opens up since the coefficient of z^2z2 is positive.
  • Further, its roots are z = 4z=4 and z = -3z=−3 which means that the value of the function is 00 for those zz values.
  • This means that the function has positive values if z> 4z>4 or z< -3z<−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>0 inequality, the part we are really interested in is the one above the xx-axis:

So, the solution is

z < -3z<−3 or z > 4z>4

or, depending on your notation preference,

z in (- oo; -3) uu (4; oo)z∈(−∞;−3)∪(4;∞).

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