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### Inequalities

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### Absolute value problems | Radical inequality problems

#### Absolute value problems

[4000] Solve: |x - 3| < 2x + 5

**Solution:** The 2 inequalities we have to solve are:

x - 3 < 2x + 5

x - 3 > -(2x + 5)

The solution to the problem has to include the solution sets from both inequalities. The solution to the first inequality:

x - 3 < 2x + 5

-8 < x

So the solution to the first inequality is x > -8. Now for the second inequality:

x - 3 > -(2x + 5)

x - 3 > -2x - 5

3x > -2

x > -2/3

So the solution to the second inequality is x > -2/3. Since the final solution has to include both solutions, our final solution is x > -2/3.

[4001] Solve: |5x + 1| > 3x - 2**Solution:** The 2 inequalities we have to solve are:

5x + 1 > 3x - 2

5x + 1 < -(3x - 2)

The solution to the problem has to include the solution sets from both inequalities. The solution to the first inequality:

5x + 1 > 3x - 2

2x > -3

x > -3/2

So the solution to the first inequality is x > -3/2. Now for the second inequality:

5x + 1 < -(3x - 2)

5x + 1 < -3x + 2

8x < 1

x < 1/8

So the solution to the second inequality is x < 1/8. Since the final solution has to include both solutions, our final solution is -3/2 < x < 1/8.

[4002] Solve: |x + 2| < x^{2} + 3x + 2

**Solution:** The 2 inequalities we have to solve are:

x + 2 < x^{2} + 3x + 2

x + 2 > -(x^{2} + 3x + 2)

The solution to the problem has to include the solution sets from both inequalities. The solution to the first inequality:

x + 2 < x^{2} + 3x + 2

0 < x^{2} + 2x

0 < x(x + 2)

Here it gets a little tricky. If
x > 0 and
x + 2 > 0 then both terms multiplied
together are greater than zero. This is satisfied if
x > 0.

On the other hand, x(x + 2) > 0 is also
satisfied if both x < 0 and
x + 2 < 0 are satisfied, which happens
when x < -2. So, the final solution for
the first inequality is x > 0 or
x < -2. Now for the second
inequality:

x + 2 > -(x^{2} + 3x + 2)

x + 2 > -x^{2} - 3x - 2

x^{2} + 4x + 4 > 0

(x + 2)^{2} > 0

Since any number other than -2 satisfies the above inequality, the
solution is the **set of real numbers except for -2.**
Since the final solution has to include both solutions, our final solution
set is x > 0 or
x < -2.

#### Radical inequality problems

[5000] Solve: correct to 4 decimal places.

First of all, since we can't have the square root of a negative
number, we have the condition

As well, since the left hand side is greater than 0, and the left
hand side is less than the right hand side, that means **the right hand
side must be greater than 0 as well.** That means we have the
condition of x > 0. Combining these two
conditions together, we have the condition
x > 0.

Now that we have that out of the way, we can square both sides:

3x + 2 < 16x^{2}

0 < 16x^{2} - 3x - 2

Since this is a parabola, we are looking for the parts of the
parabola greater than 0. Using the
quadratic formula
to find the roots, we find
the roots are -0.2720 and
0.4595. Since this parabola opens upward
(because the coefficient
of the x^{2} term is positive), the solution to our inequality is
x < -0.2720 or
x > 0.4595.

But, if you recall from the beginning of the problem, we have to satisfy the condition of x > 0. So, the only solution to our problem is x > 0.4595.

[5001] Solve:

The first thing we have to get out of the way is that we can't have the square root of a negative number. That means 2x + 2 > 0 or x > -1.

There are two cases we have to consider:

3 - x < 0

3 - x ³ 0

Let's do the easier case of 3 - x < 0. Obviously, this is the same as x > 3. Since the left hand side is greater than 0 and the right hand side is less than zero when x > 3, the solution for the first case is x > 3. The geniuses among you will have noticed that this automatically takes care of the square root condition stated above.

To solve the second case, we square both sides:

2x + 2 > (3 - x)^{2}

2x + 2 > x^{2} - 6x + 9

x^{2} - 8x + 7 < 0

(x - 1)(x - 7) < 0

Now, the only way the last inequality can be satisfied is if x - 1 > 0 and x - 7 < 0 so that the product of the 2 terms are less than zero. The solution set for this is 1 < x < 7.

But, we're not done yet. Remember the right hand side? We have 3 - x ³ 0 or x £ 3. Combining the two solutions together we get 1 < x £ 3. Finally, combining the solutions for both cases together, we come up with the final solution of x > 1.

[5002] Solve:

This problem has more parts to it than Frankenstein had. You know the old joke, "How do you eat an elephant? One bite at a time." That's how we'll tackle this problem.

Before we do anything else, we have to take care of the condition that you can't have the square root of a negative number. That means 5x - 7 > 0 or x > 1.4.

Now that we have that cleared away, the first step is getting rid of the absolute value. Doing that, we have 2 inequalities to solve:

Just so we don't get too confused, let's refer to the first inequality as case A and the second as case B. Each case will have subcases, which we will refer to as case A1 or case A2 and so on. In the end, we will combine all the cases to come up with one final solution. Are you ready?

**Case A:** I hate working with negative signs in front of a
radical, so the first thing I'm going to do is multiply both sides by

We have 2 subcases: case A1 which is 2 - x < 0 and case A2 which is 2 - x ³ 0.

Case A1 is the easy case because if the right hand side is less than 0, the left hand side is automatically greater than the right hand side since the left hand side is greater than 0. The solution for case A1 is x > 2. Note that this takes of the square root condition at the beginning of the problem.

Now for case A2. The right hand side is the same as x £ 2. But, because of the square root condition at the beginning of the problem, the solution set is 1.4 < x £ 2.

With that in mind we can now square both sides:

5x - 7 > x^{2} - 4x + 4

x^{2} - 9x + 11 < 0

In order to solve this inequality, we have to find the roots using the
quadratic formula. We
find the roots are 1.4586 and
7.5414. Since this is a parabola that opens
upward (because the
coefficient
of the x^{2} term is positive), the solution to our inequality is
1.4586 < x < 7.5414. However, once we
combine it with the solution set from the right hand side, the solution
for case A2 is
1.4586 < x £ 2.
Finally, combining the solution for both subcases A1 and A2, the
solution for case A is x > 1.4586.

**Case B:** Just like case A, we have 2 subcases: case B1 which is
x - 2 < 0 and case B2 which is
x - 2 ³ 0.

As before, we'll tackle case B1 first. The solution to this is
x < 2. Because of the square root
condition, our solution for this subcase is
1.4 < x < 2.

For subcase B2 the condition is
x ³ 2. When
we square both sides, we end up with the same inequality as in Case A
with the solution 1.4586 < x < 7.5414.
Combining the two solutions together, we get
2 £ x < 7.5414.
When we combine the solutions for subcase B1 and subcase B2, we get
1.4 < x < 7.5414.

Finally, when we combine the solution for Case A:

x > 1.4586

with the solution for Case B:

1.4 < x < 7.5414.

we come up with the final solution 1.4586 < x < 7.5414.## Algebra Help

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