Laura B.

asked • 08/28/19# Solve the question algebraically

|4x-3| = 5 square root (x+4)

note that 4x-3 is in between absolute value bars and x+4 is being square rooted completely, not just x

i know the answer is supposed to be -1.3029 and 4.3654 but I do not know how to get there algebraically.

Please help.

## 2 Answers By Expert Tutors

Jayson K. answered • 08/28/19

Mr. K's Tutoring

Whenever you have absolute values you have to break this into cases: one where you take the positive version, the second when you take the negative version.

|4x - 3| = 5√(x + 4)

Case 1: When 4x - 3 ≥ 0 → x ≥ 3/4

4x - 3 = 5√(x + 4)

(4x - 3)^{2} = [5√(x + 4)]^{2}

16x^{2} - 24x + 9 = 25(x + 4)

16x^{2} - 24x + 9 = 25x + 100

16x^{2} - 49x - 91 = 0

Using quadratic formula

x ≈ -1.3029, 4.3654

However, recall that x ≥ 3/4, so x ≈ 4.3654 is the only solution that fits in this case.

Case 2: 4x - 3 < 0 → x < 3/4

-(4x - 3) = 5√(x + 4) <-------------It's negative 4x - 3 because absolute values are suppose to keep everything positive. So if 4x - 3 always is a negative number, you have to negate (4x - 3) in order to make it positive (without the absolute value sign).

Solve the same way as above

16x^{2} - 49x - 91 = 0

x ≈ -1.3029, 4.3654

Of course, this time your answer must fit the condition that x < 3/4, which means that x ≈ -1.3029 is the only answer that works in this situation.

Thus, your answers are x ≈ -1.3209, 4.3654

Note: The whole idea is to get rid of the absolute value sign, b/c it's painful to work with them. Thus, breaking them up into separate cases, (by saying 4x - 3 is positive or negative, makes the process easier). When 4x - 3 is positive (case 1) then the absolute values don't need to be there, because the number is already positive. When 4x - 3 is negative (case 2) you have to negate 4x - 3 in order for the value to become positive (if you are to remove the absolute value signs).

Absolute values are a tricky thing,

Hope this helps

Mr. K

Laura B.

Thank you so much! I appreciate it greatly!08/28/19

Matthew S. answered • 08/28/19

University of Pittsburgh Graduate for Math Tutoring

The best place to start is to square both sides of the equation to cancel the square root. So (5sqrt(x+4))^{2} = 25(x+4).

You may be wondering: What happens to |4x-3| when you square it? The function |4x-3| is always positive. It says that when 4x-3 >= 0 then |4x-3| = 4x-3.

It also says that when 4x-3 < 0 then |4x-3| = -(4x-3), so this negative sign makes the value positive again.

When you square both (4x-3) and -(4x-3), they are the same because

[-(4x-3)]^{2} = [(-1)(4x-3)]^{2 }= (-1)^{2 }(4x-3)^{2} = (4x-3)^{2}. So |4x-3|^{2} = (4x-3)^{2}.

From here you want to use algebra to get a quadratic equation on one side and a zero on the other side,

(like ax^{2} + bx + c = 0), so you can find the roots of this function using the quadratic formula.

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Michael S.

Because there is an absolute value, you need to set up 2 equations. 1. (4x-3) = 5 sqrt(x+4) 2. -(4x-3) = 5 sqrt(x+4) In this case, it won't matter because you're going to square both sides to get rid of the root. 1a. (4x-3)^2 = (5 *sqrt(x+4))^2 1b. 16x^2 - 24x + 9 = 25 (x+4) 1c. 16x^2 - 49x - 91 = 0 Same path on 2 will give you the same quadratic equation as 1c.08/28/19