How do you solve #x( x - 5) > 6#?

1 Answer
Jacobi J.
Jul 4, 2018

#x>6#, #x< -1#

Explanation:

Let's distribute the #x# on the left side to get

#x^2-5x>6#

We have a second-degree term, so this is beginning to look like a quadratic. To put it in a more traditional form, let's subtract #6# from both sides.

#x^2-5x-6>0#

Let's see if we can factor the left. What two numbers sum up to #-5# and have a product of #-6#?

After some trial and error, we arrive at #-6# and #1#. This allows us to factor this as

#(x-6)(x+1)>0#

We have two scenarios:

#x-6>0# and #x+1>0#

We can solve them simultaneously to get

#x>6# and #x> -1#

We essentially have two conflicting scenarios. Let's think for a minute. We had the following:

#(x-6)(x+1)>0#

If the product of two things is greater than zero (positive), they must have the same signs.

#x-6# is negative if #x<6#, and positive if #x>6#.

#x+1# is negative if #x<-1#, and positive if #x> -1#

When is #(x-6)(x+1)# positive?

If #x=-1# or #6#, we will be equal to zero, but we are positive when #x>6# and if #x< -1#. These are our solutions.

If the latter part of my solution seemed confusing, what I did is essentially a sign chart. More information can be found here.

Hope this helps!

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