Exercise 67 Page 493

Practice makes perfect

a We want to solve the given quadratic equation. To do this, we will take the square root of both sides of the given equation. Since this method gives two solutions — a negative and a positive solution — remember to consider them both by adding ± to the solution.

(x+4)^2=49

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x+4)^2)=sqrt(49)

CalcRoot

Calculate root

x+4=± 7

SubEqn

LHS-4=RHS-4

x=- 4 ± 7

The solutions for this equation are x=- 4 ± 7. Let's separate them into the positive and negative cases.

x=- 4 ± 7
x_1=- 4 + 7	x_2=- 4 - 7
x_1=3	x_2=- 11

Taking the square roots, we found that the solutions of the given equation are x_1=3 and x_2=- 11.

Checking Our Answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=3.

(x+4)^2=49

Substitute

x= 3

( 3+4)^2? =49

▼

Simplify

AddTerms

Add terms

(7)^2? =49

CalcPow

Calculate power

49=49 ✓

Substituting and simplifying created a true statement, so we know that x=3 is a solution of the equation. Let's move on to x=- 11.

(x+4)^2=49

Substitute

x= - 11

( - 11+4)^2? =49

▼

Simplify

AddTerms

Add terms

(- 7)^2? =49

NegBaseToPosPow

(- a)^2 = a^2

49=49 ✓

Again, we created a true statement. x=- 11 is indeed a solution of the equation.

b To solve the given equation, let's isolate the root on one side of the given equation first. We will do it by dividing both sides of the equation by 3.

3 sqrt(x+2)=12

DivEqn

.LHS /3.=.RHS /3.

sqrt(x+2)=4

Now, to solve the above equation we can raise both sides of the given equation to the power of 2.

sqrt(x+2)=4

RaiseEqn

LHS^2=RHS^2

( sqrt(x+2) )^2=4^2

PowSqrt

( sqrt(a) )^2 = a

x+2=4^2

CalcPow

Calculate power

x+2=16

SubEqn

LHS-2=RHS-2

x=14

Raising both sides of the equation to the power of 2, we found that the solution of the given equation is x=14. We can substitute our solution back into the given equation and simplify to check if our answer is correct.

3 sqrt(x+2)=12

Substitute

x= 14

3 sqrt(14+2)? =12

▼

Simplify

AddTerms

Add terms

3sqrt(16)? =12

CalcRoot

Calculate root

3(4)? =12

Multiply

12=12 ✓

Substituting and simplifying created a true statement, so we know that x=14 is a solution of the equation.

c To solve the given equation we should start by isolating the x-term on the left-hand side of the equation.

2/x+3/10=13/10

SubEqn

LHS-3/10=RHS-3/10

2/x=10/10

Now, to simplify the equation with fractions on both sides of the equal sign we can use the Cross Product Property.

2/x=10/10

CrossMult

Cross multiply

x * 10 = 2 * 10

DivEqn

.LHS /10.=.RHS /10.

x=2

The solution to the equation is x=2. We can check our solution by substituting it into the original equation.

2/x+3/10=13/10

Substitute

x= 2

2/2+3/10 ? = 13/10

▼

Simplify

ExpandFrac

a/b=a * 5/b * 5

10/10+3/10 ? = 13/10

AddFrac

Add fractions

13/10=13/10 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

d To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left-hand sides of the equation.

5(2x-1)-2=13

Distr

Distribute 5

10x-5-2=13

Now we can continue to solve using the Properties of Equality.

10x-5-2=13

SubTerm

Subtract term

10x-7=13

AddEqn

LHS+7=RHS+7

10x=20