a. 2x = sqr root 12x+72 b. sqr root x+5 = 5 - sqr root x c. |2x| = -|x+6| d. 5 = |x+4|+|x-1| e. |x-2| = 4 -|x-3|Please Help!!!

a) Supposing that you want to
solve:

2x = sqr root
(12x+72)

First you have to check for what x values, the
square roots exists. To check, you have to solve the
inequality:

12x+72>=0

If
you divide the inequality with 4, you'll
have:

3x+18>=0

3x>=-18

x>=-18/3

x>=-6

That
means that, solving the equation, you have to keep all x values which belong to the
interval set by the condition of existence of the square
root.

Now, let's solve
it:

(2x)^2=(sqr root
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

Though
both solutions belongs to the interval, we have to verify them into the
equation.

If we put x=-3 in equation, we'll
have:

-6=sqrt(-36+72)

-6=sqrt36

-6=6,
which is not true.

b) sqr root (x+5) = 5 - sqr
root x

x+5 = 25-10sqrt
x+x

20-10sqrt x=0

2-sqrt
x=0

sqrt x =
2

x=4

c)
|2x| = -|x+6|

In order to solve this equation, first let's
see for what value of x, 2x>=0

|2x|=2x for
x>=0

|2x|=-2x, for
x<0.

|x+6|=x+6, for
x>=-6

|x+6|=-x-6, for
x<6

From these condition, occure 3
cases:

1) x belongs to (-inf.,
-6)

-2x=-(-x-6)

-2x=x+6

-3x=6

x=-2
which is not in the interval (-inf., -6).

2) x belongs to
[-6,0)

-2x=-(x+6)

-2x+x=-6

-x=-6

x=6
which belongs to the interval [-6,0).

3) x belongs to
[0,+inf.)

2x=-x-6

3x=-6

x=-2,
which is not i the interval [0,+inf.).

The
only solution of the equation is
x=6.

d)  5= |x+4|+|x-1|

In
order to solve this equation, first let's see for what value of x,
|x+4|>0

|x+4|=x+4, for
x>=-4

|x+4|=-x-4, for
x<-4

Now, let's see, for what values of
x,|x-1|>0.

|x-1|=x-1, for
x>=1

|x-1|=-x+1, for
x<1

From these condition, occure 3
cases:

1)x belongs to (-inf.,
-4)

5=-x-4-x+1

8=-2x

x=-4
which is not in the interval.

2) x belongs to
[-4,1)

5=x+4-x+1

5=5,
for any value from [-4,1).

3) x belongs to
[1,+inf.)

5=x+4+x-1

2=2x

x=1,
which belongs to [1, +inf.)

e) |x-2| = 4
-|x-3|

In order to solve this equation, first let's see for
what value of x,  |x-2|>0

 |x-2|=x-2, for
x>=2

 |x-2|=-x+2, for
x<2

Now, let's see, for what values of
x,|x-3|>0.

|x-3|=x-3, for
x>=3

|x-3|=-x+3, for
x<3

From these condition, occure 3
cases:

1) x belongs to (-inf.,
2)

-x+2=4+x-3

2x=1

x=1/2
which belongs to  (-inf., 2).

2)x belongs to
[2,3)

x-2=4+x-3

-2=1, is not
true!

3) x belongs to [3,
+inf.)

x-2=4-x+3

2x=9

x=4.5which
belongs to [3, +inf.)

The solutions of the equation
are:

x={0.5,4.5}