# 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!!!

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### 2 Answers

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}**

a)

2x = sqrt12x+72. Hope sqrt 12x means (sqrt12)x and not sqrt(12x). So,

2x-sqr12 x = 72. Or

(2-sqrt12)x = 72 Or

x = 72/(2-sqrt12) = 72(2+sqrt12)/(2-12) , after rationalising the denominator,

= 72(2+2sqrt3)/(-10)

=-7.2(2+2sqrt3)

b) sqrtx+5 = 5-sqrtx. Sqrtx+5 and sqrt(x+5) are different. Hope you do not mean sqrt(x+5) on LHS.

sqrtx=-sqrtx . Adding sqrtx to both sides,

2sqrtx = 0.

sqrtx = 0 . Or x = 0.

c)

|2x| = -|x+6|

Case (i) x<-6

|2x| = -(6-x) when x<-6 Or

-2x = 6-x. Or

-2x = 6. Or

-2x+x =-6.

-x = -6

So x = 6 and x<-6 which is a contradiction.

case(2) when 6< x <0

-2x = -(x+6). Or

-2x+x = -6. Or

-x = -6. Or x = 6 and x<0 is a contradiction.

Case (iii) x>0

2x = -(x+6) Or

2x+x = -6. Or

3x=-6 Or x =-2 and x>0 is a contradiction.

So there is no solution

d) 5 = |x+4|+|x-1|

When x>=1, 5 = x+4+x-1 Or 5 = 2x+3. Or x= (5-3)/2 = 1 is a contradiction as x>=1 and x=1. So **x =1** is a solution.

When 0 = < x < 1, 5 = x+4 +(1-x). Or

5=5+2x . Or **x = 0**. Is a solution.

When -4 =< x<0, 5 = x+4+1-x, 5 =5 is an identity.So **-4<=x< 0** is a solution.

When x <4, 5 = 4-x+(1-x) . Or 5 = 5-2x Or x=0 is a contradiction as x<-4 and x=5/2 are inconsistent.

So x=1 or (-4<x<=0) are the solution

e)

|x-2| = 4-|x-3|

case (i) x>=3

x-2 =4-(x-3). Or 2x = 4+3+2 or x =9/2 is consistent with x>=3. So x= 4.5 is a solution.

Case(ii) 2<x<3

x-2 = 4-(3-x) . Or x -2 = 3+x Or -2 = 3 is inconsistent result. So 2<x<3 cannot be a solution.

When x<2,

2-x =4-(3-x). Or

2-x=4+x-3. Or

2-4+3 = 2x Or

1 = 2x . Or x = 1/2 is consistent with x<2. So x=1/2 is a solution.