Exercise 31 Page 177

Practice makes perfect

a Let's first isolate the quadratic expression in the given equation.

2(x+3)^2-5=-5

AddEqn

LHS+5=RHS+5

2(x+3)^2=0

DivEqn

.LHS /2.=.RHS /2.

(x+3)^2=0

This equation tells us that that a quadratic expression is equal to 0. This means the expression inside the parentheses must also equal 0. (x+3)^2=0 ⇒ x+3=0 ⇒ x=-3 Therefore, the solution to this equation is -3.

b Let's first isolate the quadratic expression in the given equation.

3(x-2)^2+6=9

SubEqn

LHS-6=RHS-6

3(x-2)^2=3

DivEqn

.LHS /3.=.RHS /3.

(x-2)^2=1

Now we can solve for x by taking the square root of an equation.

(x-2)^2=1

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-2)^2)=sqrt(1)

SqrtToAbs

sqrt(a^2)=|a|

|x-2|=sqrt(1)

CalcRoot

Calculate root

|x-2|=1

An absolute value measures an expression's distance from a midpoint on a number line. |x-2|= 1 This equation means that the distance is 1, either in the positive direction or the negative direction. |x-2|= 1 ⇒ lx-2= 1 x-2= -1 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|x-2|=1

lc x-2 ≥ 0:x-2 = 1 & (I) x-2 < 0:x-2 = - 1 & (II)

lcx-2=1 & (I) x-2=-1 & (II)

(I), (II): LHS+2=RHS+2

lx_1=3 x_2=1

Both 3 and 1 are solutions to the absolute value equation.

c Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

|2x-5|-6=15

AddEqn

LHS+6=RHS+6

|2x-5|=21

An absolute value measures an expression's distance from a midpoint on a number line. |2x-5|= 21 This equation means that the distance is 21, either in the positive direction or the negative direction. |2x-5|= 21 ⇒ l2x-5= 21 2x-5= -21 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|2x-5|=21

lc 2x-5 ≥ 0:2x-5 = 21 & (I) 2x-5 < 0:2x-5 = - 21 & (II)

lc2x-5=21 & (I) 2x-5=-21 & (II)

(I), (II): LHS+5=RHS+5

l2x=26 2x=-16

(I), (II):.LHS /2.=.RHS /2.

lx_1=13 x_2=-8

Both 13 and -8 are solutions to the absolute value equation.

d Let's first isolate the square root in the given equation.

3sqrt(5x-2)+1=7

SubEqn

LHS-1=RHS-1