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

1 Answer
May 16, 2018

Answer:

#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.#