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

1 Answer

Answer:

#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)#.