Solve the question algebraically

Question

|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.

Answer 1

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

16x² - 24x + 9 = 25(x + 4)

16x² - 24x + 9 = 25x + 100

16x² - 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

16x² - 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

Answer 2

When you take away the absolute value, you have 2 possible solutions. One positive version on the other side and one negative. In this case, they give the same result. So will just need to do one version.

To get rid of the square root, square both sides.

(4x-3)²=(5√x+4)² Simplify

(4x-3)(4x-3)=5²(√x+4)² Simplify using FOIL, squaring 5 and removing the radical symbol when you square it.

16x²-24x+9=25(x+4) Distributive property

16x²-24x+9=25x+100 Subtract 25x and 100 from both sides

16x²-24x+9-25x-100=25x+100-25x-100 Simplify

16x²-49x-91 = 0 Solve using the quadratic formula

a=16, b=-49 and c=-91

x=-1.3029 and 4.3654

Answer 3

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 ax² + bx + c = 0), so you can find the roots of this function using the quadratic formula.