Is the solution to the quadratic inequality (-x + 6)/(x - 2) < 0?

[mathdad]

Solve the quadratic inequality.

2x/(x - 2) < 3

Multiply both sides by (x - 2).

[(x - 2)][2x/(x - 2)] < 3(x - 2)

2x < 3x - 6

2x - 3x < -6

-x < -6

x > 6

Our only end point is x = 6.

<----------(6)---------->

For (-infinity, 6), let x = 0. In this interval, we get false.

For (6, infinity), let x = 7. In this interval, we get true.

Test x = 6.

2(6)/(6 - 2) < 3

12/4 < 3

3 < 3...false statement. We exclude x = 6 as part of the solution.

Solution:

(6, infinity)

Correct?

 

[MarkFL]

You don't want to multiply an inequality by an expression whose sign is unknown...arrange everything to one side and then get your critical values from the roots of the numerator and denominator. :D

 

[mathdad]

Are you saying the correct set up is

2x/(x - 2) - 3 < 0?

 

[MarkFL]

Are you saying the correct set up is

2x/(x - 2) - 3 < 0?

Yes, now combine terms on the LHS...:D

 

[mathdad]

Cool. I will work on this quadratic inequality later. I sure wish I had a better understanding of mathematics.

 

[mathdad]

2x/(x - 2) - 3 < 0

After combining, I got the following:

(-x + 6)/(x - 2) < 0

-x + 6 = 0

x = 6

x - 2 = 0

x = 2

<--------(2)--------(6)---------->

When x = 6, we get 0/(x - 2) < 0.

When x = 2, we get undefined.

We must exclude 2 and 6.

For (-infinity, 2), let x = 0. Here we get a true statement.

For (2, 6), let x = 3. Here we get a false statement.

For (6, infinity), let x = 7. Here we get a true statement.

Solution: (-infinity, 2) U (6, infinity)

Correct?