MHB Yes, the correct solution would be to subtract (x/2) from both sides, not 2x.

[mathdad]

Solve the quadratic inequality.

2/x < x/2

Multiply both sides by 2x.

(2x)*(2/x) < (2x)(x/2)

4 < x^2

4 = x^2

sqrt{4} = sqrt{x^2}

-2 = x

2 = x

Our end points are -2 and 2.

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

For (-infinity, -2), let x = -3. In this interval, we get true.

For (-2, 2), let x = 0. In this interval, we get false.

For (2, infinity), let x = 3. In this interval, we get true.

Test the end points.

Let x = -2 and x = 2.

At x = -2, we get false.

At x = 2, we get false.

We exclude the test points.

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

Correct?

 

[MarkFL]

Solve the quadratic inequality.

2/x < x/2

Multiply both sides by 2x.

You are assuming x is positive when you do that. A better step is to subtract x/2 from both sides:

$$\frac{2}{x}-\frac{x}{2}<0$$

Combine terms:

$$\frac{4-x^2}{2x}<0$$

$$\frac{(2+x)(2-x)}{2x}<0$$

Now we see we have 3 critical values which give us 4 intervals:

$$(-\infty,-2)$$ Test value: x = -3: expression is (-)(+)/(-) = + not part of solution. Other intervals will alternate...

$$(-2,0)$$ is part of solution.

$$(0,2)$$ not part of solution.

$$(2,\infty)$$ is part of solution.

And so the solution is:

$$(-2,0)\,\cup\,(2,\infty)$$

 

[mathdad]

You meant to say subtract (x/2) from both sides not 2x.