Exercise 8 Page 171

Practice makes perfect

To solve the quadratic equation, the first step is to use inverse operations to get the square alone on the left-hand side.

2(x-1)^2+7=39

SubEqn

LHS-7=RHS-7

2(x-1)^2=32

DivEqn

.LHS /2.=.RHS /2.

(x-1)^2=16

We can now take the square root of both sides to eliminate the power. Remember that there will be both a positive and a negative solution to the square root.

(x-1)^2=16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x+1=±sqrt(16)

CalcRoot

Calculate root

x-1=±4

AddEqn

LHS+1=RHS+1

x=1±4

StateSol

State solutions

lcx=1-4 & (I) x=1+4 & (II)

(I), (II): Add and subtract terms

lx_1=-3 x_2=5

Like in Part A, we have to perform inverse operations until the square root is by itself.

7(sqrt(m+1)-3)=21

DivEqn

.LHS /7.=.RHS /7.

sqrt(m+1)-3=3

AddEqn

LHS+3=RHS+3

sqrt(m+1)=6

Since squaring is the inverse operation to taking the square root, we have to square both sides to continue solving the equation.

sqrt(m+1)=6

RaiseEqn

LHS^2=RHS^2

m+1=36

SubEqn

LHS-1=RHS-1

m=35

When we square a function, we may introduce extraneous solutions. Therefore, our solution must be tested in the original equation to make sure its correct. |c|c|c| [-1em] -3pt m -3pt & -3pt 7(sqrt(m+1)-3)=21 -3pt & -5pt Evaluate -4pt [0.2em] [-1em] -3pt 35 -3pt & -3pt 7(sqrt(35+1)-3)? =21 -3pt & -5pt 21=21 ✓ -4pt [0.2em]

To solve the equation, we want to get rid of the fractions with multiplying both sides by the greatest denominator.

x/2+x/3=5x+2/6

MultEqn

LHS * 6=RHS* 6

6(x/2+x/3)=5x+2

Distr

Distribute 6

6*x/2+6*x/3=5x+2

MoveLeftFacToNum

a*b/c= a* b/c

6x/2+6x/3=5x+2

SimpQuot