Exercise 120 Page 303

Practice makes perfect

a An absolute value measures an expression's distance from a midpoint on a number line.

|4x+20|= 8This equation means that the distance is 8, either in the positive direction or the negative direction. |4x+20|= 3 ⇒ l4x+20= 8 4x+20= - 8 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|4x+20|=8

lc 4x+20 ≥ 0:4x+20 = 8 & (I) 4x+20 < 0:4x+20 = - 8 & (II)

lc4x+20=- 8 & (I) 4x+20=8 & (II)

(I), (II): LHS-20=RHS-20

l4x=- 28 4x=-12

(I), (II): .LHS /4.=.RHS /4.

lx_1=- 7 x_2=-3

Both -7 and -3 are solutions to the absolute value equation.

b Let's solve the equation by performing inverse operations until x is isolated. Notice that 8 is a perfect cube.

(x-13)^3=8

RadicalEqn

sqrt(LHS)=sqrt(RHS)

x-13=2

AddEqn

LHS+13=RHS+13

x=15

c Let's solve the equation by performing inverse operations until x is isolated.

2sqrt(x-4)=14

DivEqn

.LHS /2.=.RHS /2.

sqrt(x-4)=7

RaiseEqn

LHS^2=RHS^2

x-4=49

AddEqn

LHS+4=RHS+4

x=53

d Like in Part A, we have an absolute value equation. This means we get two cases when removing the absolute value. Before we can do that, we have to isolate the absolute value expression.

6|x-8|-4=14

AddEqn

LHS+4=RHS+4

6|x-8|=18

DivEqn

.LHS /6.=.RHS /6.

|x-8|=3

Now we can remove the absolute value, which gives us two cases — one where the right-hand side is positive another where it is negative.

|x-8|=3

lc 4x+20 ≥ 0:4x+20 = 8 & (I) 4x+20 < 0:4x+20 = - 8 & (II)

lcx-8=- 3 & (I) x-8=3 & (II)

(I), (II): LHS+8=RHS+8

lx_1=5 x_2=11

Both 5 and 11 are solutions to the absolute value equation.

e Like in Part B, we will perform inverse operations until x is isolated.

3(x+12)^2=27

DivEqn

.LHS /3.=.RHS /3.

(x+12)^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x+12=± 3

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Solve for x

SubEqn

LHS-12=RHS-12

x=± 3-12

StateSol

State solutions

lcx=-3-12 & (I) x=3-12 & (II)

(I), (II): Subtract term

lx_1=-15 x_2=-9

f Notice that 36 can be rewritten as a power with a base of 6. With this information, we can rewrite the equation so that we may equate the exponents and then solve for x.

6^4=36^x

WritePow

Write as a power

6^4=(6^2)^x

PowPow

(a^m)^n=a^(m* n)

6^4=6^(2x)

EquateExponents

Equate exponents

4=2x

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Solve for x