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Core Connections Algebra 1, 2013 View details

2. Section 11.2

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Exercise 57 Page 545

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a We will solve the given equation by taking the square roots. To do this we need to consider the positive and negative solutions.

x^2 - 1 = 15

AddEqn

LHS+1=RHS+1

x^2 = 16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x^2)=sqrt(16)

CalcRoot

Calculate root

x=± 4

We found that x=± 4. Thus there are two solutions of the given equation, which are x=4 and x=- 4.

b We want to solve the given system of equations using substitution, so we need to follow three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable. For this exercise y is already isolated in both equations, so we can skip straight to solving! Since the expression equal to y in (I) is simpler, let's use that for our initial substitution.
y=3x-2 & (I) y=4x+3 & (II)
Substitute
(II):y= 3x-2
y=3x-2 3x-2=4x+3
AddEqn
(II):LHS+2=RHS+2
y=3x-2 3x=4x+5
SubEqn
(II):LHS-4x=RHS-4x
y=3x-2 - x=5
DivEqn
(II): .LHS /(- 1).=.RHS /(- 1).
y=3x-2 x=- 5
Great! Now, to find the value of y we need to substitute x=- 5 into either one of the equations in the given system. Let's use the first equation.
y=3x-2 x=- 5
Substitute
(I):x= - 5
y=3( - 5)-2 x=- 5
MultPosNeg
(I): a(- b)=- a * b
y=- 15-2 x=- 5
SubTerm
(I): Subtract term
y=- 17 x=- 5
The solution, or point of intersection, to this system of equations is the point (- 5,- 17).

c We will use the Quadratic Formula to solve the given quadratic equation.

ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aWe first need to identify the values of a, b, and c. x^2-2x-8=0 ⇕ 1x^2+( - 2)x+( - 8)=0 We see that a= 1, b= - 2, and c= - 8. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -2)±sqrt(( - 2)^2-4( 1)( - 8))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(1)(- 8))/2(1)

CalcPow

Calculate power

x=2±sqrt(4-4(1)(- 8))/2(1)

IdPropMult

Identity Property of Multiplication

x=2±sqrt(4-4(- 8))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+32)/2

AddTerms

Add terms

x=2±sqrt(36)/2

CalcRoot

Calculate root

x=2± 6/2

CalcQuot

Calculate quotient

x=1 ± 3

The solutions for this equation are x=1 ± 3. Let's separate them into the positive and negative cases.

x=1 ± 3
x_1=1 + 3	x_2=1 - 3
x_1=4	x_2=- 2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=4 and x_2=- 2.

d We will use the Quadratic Formula to solve the given quadratic equation.

ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aLet's start by rewriting the equation so all of the terms are on the left-hand side.

2x^2=- x+7

AddEqn

LHS+x=RHS+x

2x^2+x=7

SubEqn

LHS-7=RHS-7

2x^2+x-7=0

Now we can identify the values of a, b, and c. 2x^2+x-7=0 ⇕ 2x^2+ 1x+( - 7)=0 We see that a= 2, b= 1, and c= - 7. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 2)( - 7))/2( 2)

▼

Solve for x and Simplify

BaseOne

1^a=1

x=- 1±sqrt(1-4(2)(- 7))/2(2)

Multiply

x=- 1±sqrt(1-8(- 7))/4

MultNegNegOnePar

- a(- b)=a* b

x=- 1±sqrt(1+56)/4

AddTerms

Add terms

x=- 1±sqrt(57)/4

Using the Quadratic Formula, we found that the solutions of the given equation are x= - 1± sqrt(57)4. Therefore, the solutions are x_1= - 1+sqrt(57)4 and x_2= - 1-sqrt(57)4.

Subchapter links

11.2.1 Investigating Data Representations p.536-538

11.2.2 Comparing Data p.543-545

11.2.3 Standard Deviation p.549-550

Exercise 57 Page 545

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