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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

1. Section 4.1

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Exercise 24 Page 176

Practice makes perfect

a

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

-3sqrt(2x-5)+7=-8

LHS-7=RHS-7

-3sqrt(2x-5)=-15

.LHS /(-3).=.RHS /(-3).

sqrt(2x-5)=5

Now we can raise each side of the equation to the power of 2.

sqrt(2x-5)=5

LHS^2=RHS^2

(sqrt(2x-5))^2=5^2

( sqrt(a) )^2 = a

2x-5=5^2

Calculate power

2x-5=25

LHS+5=RHS+5

2x=30

.LHS /2.=.RHS /2.

x=15

Finally, we will check our solution by substituting the value into the original equation.

-3sqrt(2x-5)+7=-8

x= 15

-3sqrt(2( 15)-5)+7? =-8

▼

Simplify left-hand side

Multiply

-3sqrt(30-5)+7? =-8

Subtract term

-3sqrt(25)+7? =-8

Calculate root

-3(5)+7? =-8

(- a)b = - ab

-15+7? =-8

Add terms

-8=-8

b

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

2|3x+4|-10=12

LHS+10=RHS+10

2|3x+4|=22

.LHS /2.=.RHS /2.

|3x+4|=11

An absolute value measures an expression's distance from a midpoint on a number line. |3x+4|= 11

This equation means that the distance is 11, either in the positive direction or the negative direction. |3x+4|= 11 ⇒ l3x+4= 11 3x+4= -11 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|3x+4|=11

lc 3x+4 ≥ 0:3x+4 = 11 & (I) 3x+4 < 0:3x+4 = - 11 & (II)

lc3x+4=11 & (I) 3x+4=-11 & (II)

(I), (II): LHS-4=RHS-4

l3x=7 3x=-15

(I), (II):.LHS /3.=.RHS /3.

lx_1= 73 x_2=-5

Both 73 and -5 are solutions to the absolute value equation. When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with x= 73.

2|3x+4|-10=12

x= 7/3

2|3( 7/3)+4|-10? =12

▼

Simplify

DenomMultFracToNumber

3 * a/3= a

2|7+4|-10? =12

Add terms

2|11|-10? =12

|11|=11

2(11)-10? =12

Multiply

22-10? =12

Subtract term

12=12

We will check x=-5 in the same way.

2|3x+4|-10=12

x= -5

2|3( -5)+4|-10? =12

▼

Simplify

a(- b)=- a * b

2|-15+4|-10? =12

Add terms

2|-11|-10? =12

|-11|=11

2(11)-10? =12

Multiply

22-10? =12

Subtract term

12=12

We see that both solutions satisfy the original equation.

Subchapter links

4.1.1 Strategies for Solving Equations p.171-172

4.1.2 Solving Equations and Systems Graphically p.176-178

4.1.3 Finding Multiple Solutions to Systems of Equations p.180-181

4.1.4 Using Systems of Equations to Solve Problems p.184