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|

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].