# 1. Solve the following equations: a. 2x=sqrt 12x+72 b. sqrt x+5=5-sqrt x c. |2x|=-|x+6| d. 5=|x+4|+|x-1| e. |x-2|=4 -|x-3|

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### 2 Answers

This is not an answer to the above problem. More of a question for giorgiana1976.

In part c) you said: "x=6 which belongs to the interval [-6, 0), so it is a solution of the equation." x=6 does NOT belong to that inteval. Since that was the only solution that you found, I think that there are no solutions.

Plus, shouldn't that have been apparent just by looking at the original equation? There is no way that a negative absolute value could be equal to a positive absolute value...

Correct me if I'm wrong.

a) For solving the first equation, we have to take into account the constraints for x values, for the existence of the sqrt.

12x+72>0

3x+18>0

x>-18/3

x>-6

Now let's solve the equation:

We'll square raise the expressions, both sides of the equal sign.

(2x)^2=[sqrt (12x+72)]^2

4x^2 = 12x+72

4x^2 -12x - 72 = 0

x^2 - 3x - 18 = 0

x1 = [3+sqrt(9+72)]/2

x1 = (3+9)/2

x1 = 6

x2 = (3-9)/2

x2 = -3

Now, we'll verify the solutions into the equation, even if they belong to the interval [-6, +inf.)

x1=6

12 = sqrt(72+72)

12=sqrt144

12=12 true

x1=6 is the solution of the equation.

For x2=-3

It is obvious that -6=sqrt(36) it's impossible!

**So, the equation will have just one solution, x=6!**

b) Again, for solving the equation, we have to take into account the constraints for x values, for the existence of the sqrt.

x>0 and x>-5

x belongs to the interval [0,+inf.)

We'll square raise the expressions, both sides of the equal sign.

[sqrt (x+5)]^2=(5-sqrt x)^2

x+5=25-10sqrt x + x

5-25=-10sqrt x

-20 = -10sqrt x

2 = sqrt x

**x = 4**

c) For solving |2x|=-|x+6|, we have to follow the rule of the absolute value.

For x belongs to the interval (-inf, -6)

-2x=-(-x-6)

-2x-x=6

-3x=6

x=-2 doesn't belong to the interval (-inf, -6), so is not a solution of the equation.

For x belongs to the interval [-6, 0)

-2x=-(x+6)

-2x+x=-6

-x=-6

x=6 which belongs to the interval [-6, 0), so it is a solution of the equation.

For x belongs to the interval [0, inf)

2x=-(x+6)

2x+x=-6

3x=-6

x=-2 doesn't belong to the interval [0, inf), so is not a solution of the equation.

**So, the equation will have just one solution, x=6!**

d) For solving 5=|x+4|+|x-1|, we have to follow the rule of the absolute value.

For x belongs to the interval (-inf, -4)

5=-x-4-x+1

8=-2x

x=-4 doesn't belong to the interval (-inf, -4), so is not a solution of the equation.

For x belongs to the interval [-4, 1)

x+4-x+1=5

5=5

So, any value from the interval [-4,1) will be a solution for the equation.

For x belongs to the interval [1, inf)

x+4+x-1=5

2x=2

x=1 which belongs to the interval [1, inf), so it is a solution of the equation.

So, the solutions of the equation will be in the interval [-4,1].