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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

3. Section 11.3

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Exercise 94 Page 608

Practice makes perfect

a We want to solve the given rational equation.

x/3=4/x Let's begin by highlighting the all of the different factors in the denominators. This will help us find the least common denominator (LCD). x/3=4/x We will solve the equation by multiplying each side by the LCD to clear denominators.

x/3=4/x

LHS * (3) (x)=RHS* (3) (x)

x/3 * (3) (x)=4/x * (3) (x)

Multiply

x(3)(x)/3=4(3)(x)/x

CancelCommonFac

Cancel out common factors

x(3)(x)/3=4(3)(x)/x

Simplify quotient

x(x)=4(3)

▼

Simplify

Multiply

x^2=12

sqrt(LHS)=sqrt(RHS)

x = ± sqrt(12)

Finally, we will check if either x= sqrt(12) or x= - sqrt(12) is an extraneous solution. To do this, we need to substitute sqrt(12) and - sqrt(12) for x in the original equation. Let's start with x= sqrt(12).

x/3=4/x

x= sqrt(12)

sqrt(12)/3 ? = 4/sqrt(12)

▼

Simplify

a/b=a * sqrt(12)/b * sqrt(12)

sqrt(12)/3 ? = 4 sqrt(12)/(sqrt(12))^2

SqrtPowToNumber

sqrt(a^2)=a

sqrt(12)/3 ? = 4sqrt(12)/12

a/b=.a /4./.b /4.

sqrt(12)/3 = sqrt(12)/3 ✓

Now, we will check our second solution, x= - sqrt(12).

x/3=4/x

x= - sqrt(12)

-sqrt(12)/3 ? = 4/-sqrt(12)

▼

Simplify

a/b=a * - sqrt(12)/b * - sqrt(12)

-sqrt(12)/3 ? = 4 (-sqrt(12))/( - sqrt(12))^2

NegBaseToPosBase

(- a)^2=a^2

-sqrt(12)/3 ? = 4 (-sqrt(12))/( sqrt(12))^2

SqrtPowToNumber

sqrt(a^2)=a

- sqrt(12)/3 ? = 4(-sqrt(12))/12

a/b=.a /4./.b /4.

- sqrt(12)/3 =- sqrt(12)/3 ✓

Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are x=sqrt(12) and x=- sqrt(12).

b We want to solve the given rational equation.

x/x-1=4/x Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD. x/x-1=4/xWe will solve the equation by multiplying each side by the LCD to clear denominators.

x/x-1=4/x

LHS * (x-1) (x)=RHS* (x-1) (x)

x/x-1 * (x-1) (x)=4/x * (x-1) (x)

Multiply

x(x-1)(x)/x-1 = 4(x-1)(x)/x

CancelCommonFac

Cancel out common factors

x(x-1)(x)/x-1 = 4(x-1)(x)/x

Simplify quotient

x(x)=4(x-1)

▼

Simplify

Multiply

x^2=4(x-1)

Distribute 4

x^2=4x-4

LHS-4x=RHS-4x

x^2-4x=-4

LHS+4=RHS+4

x^2-4x+4=0

Note that we have a quadratic equation now. Let's identify the values of a, b, and c. x^2-4x+4=0 ⇔ 1x^2 +(-4)x+4=0 We have that a= 1, b=-4, and c=4. Let's substitute these values into the Quadratic Formula and solve for x.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- -4±sqrt((-4)^2-4( 1)(4))/2( 1)

▼

Simplify right-hand side

- (- a)=a

x=4±sqrt((-4)^2-4(1)(4))/2(1)

Calculate power and product

x=4±sqrt(16-16)/2

Subtract term

x=4±sqrt(0)/2

Calculate root

x=4 ± 0/2

Identity Property of Addition

x=4/2

Calculate quotient

x=2

Finally, we will check if x=2 is an extraneous solution. To do this, we need to substitute 2 for x in the original equation.

x/x-1=4/x

x= 2

2/2-1? =4/2

▼

Simplify

Subtract term

2/1? =4/2

Calculate quotient

2=2 ✓

x=2 is not an extraneous solution, so it solves our equation.

c We want to solve the given rational equation.

1/x+1/3x=6 Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD. 1/x+1/3x=6 We will solve the equation by multiplying each side by the LCD to clear denominators.

1/x+1/3x=6

LHS * ( 3x)=RHS* ( 3x)

( 1/x+1/3x) * ( 3x) =6 * ( 3x)

Multiply

3x/x + 3x/3x = 18x

CancelCommonFac

Cancel out common factors

3x/x + 3x/3x = 18x

Simplify quotient

3+1=18x

▼

Simplify

Add terms

4=18x

.LHS /18.=.RHS /18.

4/18=x

a/b=.a /2./.b /2.

2/9=x

Rearrange equation

x=2/9

Finally, we will check if x= 29 is an extraneous solution. To do this, we need to substitute 29 for x in the original equation.

1/x+1/3x=6

x= 2/9

1/.2 /9.+1/3( .2 /9.)? =6

▼

Simplify

MoveLeftFacToNum

a*b/c= a* b/c

1/.2 /9.+1/.6 /9.? =6

a/b/c= a * c/b

9/2+9/6? =6

a/b=a * 3/b * 3

27/6+9/6? =6

Add fractions

36/6? =6

Calculate quotient

6=6 ✓

x = 29 is not an extraneous solution, so it solves our equation.

d We want to solve the given rational equation.

1/x+1/x+1=3 Let's begin by highlighting the all of the different factors in the denominators. This will help us find the LCD. 1/x+1/x+1=3We will solve the equation by multiplying each side by the LCD to clear denominators.

1/x+1/x+1=3

LHS * (x) (x+1)=RHS* (x) (x+1)

(1/x+1/x+1) * (x) (x+1)=3 * (x) (x+1)

Multiply

x(x+1)/x+x(x+1)/x+1=3x(x+1)

CancelCommonFac

Cancel out common factors

x(x+1)/x+x(x+1)/x+1 = 3x(x+1)

Simplify quotient

x+1+x=3x(x+1)

▼

Simplify

Add terms

2x+1=3x(x+1)

Distribute 3x

2x+1=3x^2+3x

LHS-3x^2=RHS-3x^2

-3x^2+2x+1=3x

LHS-3x=RHS-3x

-3x^2-x+1=0

Note that we have a quadratic equation now. Let's identify the values of a, b, and c. -3x^2-x+1=0 ⇔ - 3x^2 +(-1)x+1=0 We have that a= - 3, b=-1, and c=1. Let's substitute these values into the Quadratic Formula and solve for x.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- (-1)±sqrt((-1)^2-4( - 3)(1))/2( - 3)

▼

Simplify right-hand side

- (- a)=a

x=1±sqrt((-1)^2-4(-3)(1))/2(-3)

Calculate power and product

x=1±sqrt(1+12)/- 6

Add terms

x=1±sqrt(13)/- 6

Let's calculate both solutions by using the positive and negative signs.

x=1 ± sqrt(13)/- 6
x=1 + sqrt(13)/- 6	x=1 - sqrt(13)/- 6
x=-(1 + sqrt(13))/6	x=-(1 - sqrt(13))/6
x=-1 - sqrt(13)/6	x=-1 + sqrt(13)/6

Finally, we will check if either x= -1 - sqrt(13)6 or x= -1 + sqrt(13)6 is an extraneous solution. To do this, we need to substitute -1 - sqrt(13)6 and -1 + sqrt(13)6 for x in the original equation. Let's start with x= -1 - sqrt(13)6.

1/x+1/x+1=3

x= -1 - sqrt(13)6

1/-1 - sqrt(13)6+1/-1 - sqrt(13)6+1=3

▼

Simplify

a/b=a * 6/b * 6

1/-1 - sqrt(13)6+1/-1 - sqrt(13)6+ 66=3

Add fractions

1/-1 - sqrt(13)6+1/-1 - sqrt(13)+66=3

Add terms

1/-1 - sqrt(13)6+1/- sqrt(13)+56=3

a/b/c= a * c/b

6/-1 - sqrt(13)+6/- sqrt(13)+5=3

a/b=a * (- sqrt(13)+5)/b * (- sqrt(13)+5)

6(- sqrt(13)+5)/(-1 - sqrt(13))(- sqrt(13)+5)+6/- sqrt(13)+5=3

a/b=a * (-1 - sqrt(13))/b * (-1 - sqrt(13))

6(- sqrt(13)+5)/(-1 - sqrt(13))(- sqrt(13)+5)+6(-1 - sqrt(13))/(- sqrt(13)+5)(-1 - sqrt(13))=3

Add fractions

6(- sqrt(13)+5)+6(-1 - sqrt(13))/(-1 - sqrt(13))(- sqrt(13)+5)=3

Distribute 6

- 6sqrt(13)+30-6 - 6sqrt(13)/(-1 - sqrt(13))(- sqrt(13)+5)=3

Add and subtract terms

24-12sqrt(13)/(-1 - sqrt(13))(- sqrt(13)+5)=3

Distribute (-1 - sqrt(13))

24-12sqrt(13)/- sqrt(13)(-1 - sqrt(13))+5(-1 - sqrt(13))=3

Distribute - sqrt(13)

24-12sqrt(13)/sqrt(13)+ 13+5(-1 - sqrt(13))=3

Distribute 5

24-12sqrt(13)/sqrt(13)+ 13-5 - 5sqrt(13)=3

Add and subtract terms

24-12sqrt(13)/8-4sqrt(13)=3

Factor out 3

3(8-4sqrt(13))/8-4sqrt(13)=3

a/b=.a /(8-4sqrt(13))./.b /(8-4sqrt(13)).

3/1=3

Calculate quotient

3=3 ✓

Now, we will check our second solution, x= -1 + sqrt(13)6.

1/x+1/x+1=3

x= -1 + sqrt(13)6

1/-1 + sqrt(13)6+1/-1 + sqrt(13)6+1=3

▼

Simplify

a/b=a * 6/b * 6

1/-1 + sqrt(13)6+1/-1 + sqrt(13)6+ 66=3

Add fractions

1/-1 + sqrt(13)6+1/-1 + sqrt(13)+66=3

Add terms

1/-1 + sqrt(13)6+1/sqrt(13)+56=3

a/b/c= a * c/b

6/-1 + sqrt(13)+6/sqrt(13)+5=3

a/b=a * ( sqrt(13)+5)/b * ( sqrt(13)+5)

6(sqrt(13)+5)/(-1 + sqrt(13))(sqrt(13)+5)+6/sqrt(13)+5=3

a/b=a * (-1 + sqrt(13))/b * (-1 + sqrt(13))

6( sqrt(13)+5)/(-1 + sqrt(13))( sqrt(13)+5)+6(-1 + sqrt(13))/(sqrt(13)+5)(-1 + sqrt(13))=3

Add fractions

6(sqrt(13)+5)+6(-1 + sqrt(13))/(-1 + sqrt(13))(sqrt(13)+5)=3

Distribute 6

6sqrt(13)+30-6 +6sqrt(13)/(-1 + sqrt(13))(sqrt(13)+5)=3

Add and subtract terms

24+12sqrt(13)/(-1 + sqrt(13))(sqrt(13)+5)=3

Distribute (-1 - sqrt(13))

24+12sqrt(13)/sqrt(13)(-1 + sqrt(13))+5(-1 + sqrt(13))=3

Distribute sqrt(13)

24+12sqrt(13)/- sqrt(13)+ 13+5(-1 + sqrt(13))=3

Distribute 5

24+12sqrt(13)/- sqrt(13)+ 13-5 + 5sqrt(13)=3

Add and subtract terms

24+12sqrt(13)/8+4sqrt(13)=3

Factor out 3

3(8+4sqrt(13))/8+4sqrt(13)=3

a/b=.a /(8+4sqrt(13))./.b /(8+4sqrt(13)).

3/1=3

Calculate quotient

3=3 ✓

Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are x= -1 - sqrt(13)6 and x= -1 + sqrt(13)6.

Subchapter links

11.3.1 Analyzing Decisions and Strategies p.607-608