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Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 10.2
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Exercise 92 Page 503

Practice makes perfect
a To solve the given quadratic equation, let's isolate the square on one side of the given equation first. We will do it by dividing both sides of the equation by 9.
9(x-4)^2=81
DivEqn

.LHS /9.=.RHS /9.

(x-4)^2=9
We want to solve the above equation. To do this, we will take the square root of both sides of the equation. Since this method gives two solutions — a negative and a positive — remember to consider them both by adding ± to the solution.
(x-4)^2=9
SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-4)^2)=sqrt(9)
CalcRoot

Calculate root

x-4=± 3
AddEqn

LHS+4=RHS+4

x=4 ± 3
The solutions for this equation are x=4 ± 3. Let's separate them into the positive and negative cases.
x=4 ± 3
x_1=4 + 3 x_2=4 - 3
x_1=7 x_2=1

Taking the square roots, we found that the solutions of the given equation are x_1=7 and x_2=1.

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=7.
9(x-4)^2=81
Substitute

x= 7

9( 7-4)^2? =81
â–Ľ
Simplify
SubTerm

Subtract term

9(3)^2? =81
CalcPow

Calculate power

9(9)? =81
Multiply

Multiply

81=81 âś“
Substituting and simplifying created a true statement, so we know that x=7 is a solution of the equation. Let's move on to x=1.
9(x-4)^2=81
Substitute

x= 1

9( 1-4)^2? =81
â–Ľ
Simplify
SubTerm

Subtract term

9(- 3)^2? =81
NegBaseToPosPow

(- a)^2 = a^2

9(9)? =81
Multiply

Multiply

81=81 âś“
Again, we created a true statement. x=- 11 is indeed a solution of the equation.
b An absolute value measures an expression's distance from a midpoint on a number line.
|x-6|= 2 This equation means that the distance is 2, either in the positive direction or the negative direction. |x-6|= 2 ⇒ lx-6= 2 x-6= - 2 To find the solutions to the absolute value equation we need to solve both of these cases for x.
| x-6|=2

lc x-6 ≥ 0:x-6 = 2 & (I) x-6 < 0:x-6 = - 2 & (II)

lcx-6=2 & (I) x-6=- 2 & (II)

(I), (II): LHS+6=RHS+6

lx_1=8 x_2=4
Both 8 and 4 are solutions to the absolute value equation.

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=8.
|x-6|=2
Substitute

x= 8

| 8-6|? =2
â–Ľ
Simplify
SubTerm

Subtract term

|2|? =2
AbsPos

|2|=2

2=2 âś“
Substituting and simplifying created a true statement, so we know that x=8 is a solution of the equation. Let's move on to x=4.
|x-6|=2
Substitute

x= 4

| 4-6|? =2
â–Ľ
Simplify
SubTerm

Subtract term

|- 2|? =2
AbsNeg

|-2|=2

2=2 âś“
Again, we created a true statement. x=2 is indeed a solution of the equation.
c To solve the given equation, let's first isolate the square root on one side of the equation. We will do it by substituting 2 from both sides of the equation.
5=2+sqrt(3x)
SubEqn

LHS-2=RHS-2

3=sqrt(3x)
Now, to solve the above equation, we can raise both sides of the given equation to the power of 2.
3=sqrt(3x)
RaiseEqn

LHS^2=RHS^2

3^2=( sqrt(3x) )^2
PowSqrt

( sqrt(a) )^2 = a

3^2=3x
â–Ľ
Solve for x
CalcPow

Calculate power

9=3x
DivEqn

.LHS /3.=.RHS /3.

3=x
RearrangeEqn

Rearrange equation

x=3
Raising both sides of the equation to the power of 2, we found that the solution of the given equation is x=3. We can substitute our solution back into the given equation and simplify to check if our answer is correct.
5=2+sqrt(3x)
Substitute

x= 3

5? =2+sqrt(3( 3))
â–Ľ
Simplify
Multiply

Multiply

5? =2+sqrt(9)
CalcRoot

Calculate root

5? =2+3
AddTerms

Add terms

5=5 âś“
Substituting and simplifying created a true statement, so we know that x=3 is a solution of the equation.
d Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.
2|x+1|=- 4
DivEqn

.LHS /2.=.RHS /2.

|x+1|=- 2
An absolute value measures an expression's distance from a midpoint on a number line. Since distance cannot be negative, the absolute value of a number cannot be negative. | x+1|= - 2 An absolute value cannot be equal to -2, so this absolute value equation has no solution.
Subchapter links expand_more
arrow_right 10.2.1 Solving by Rewriting p.479-480
28
Solving by Rewriting 29 (Page 480)
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Solving by Rewriting 31 (Page 480)
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Solving by Rewriting 33 (Page 480)
arrow_right 10.2.2 Fraction Busters p.483-484
Fraction Busters 39 (Page 483)
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Fraction Busters 43 (Page 484)
Fraction Busters 44 (Page 484)
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arrow_right 10.2.3 Multiple Methods for Solving Equations p.488-489
Multiple Methods for Solving Equations 53 (Page 488)
54
55
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Multiple Methods for Solving Equations 57 (Page 489)
58
Multiple Methods for Solving Equations 59 (Page 489)
arrow_right 10.2.4 Determining the Number of Solutions p.493-494
Determining the Number of Solutions 67 (Page 493)
68
Determining the Number of Solutions 69 (Page 493)
70
71
Determining the Number of Solutions 72 (Page 494)
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arrow_right 10.2.5 Deriving the Quadratic Formula and the Number System p.499-500
81
Deriving the Quadratic Formula and the Number System 82 (Page 499)
Deriving the Quadratic Formula and the Number System 83 (Page 500)
Deriving the Quadratic Formula and the Number System 84 (Page 500)
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86
87
arrow_right 10.2.6 More Solving and an Application p.502-503
90
91
More Solving and an Application 92 (Page 503)
More Solving and an Application 93 (Page 503)
More Solving and an Application 94 (Page 503)
More Solving and an Application 95 (Page 503)
More Solving and an Application 96 (Page 503)