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Core Connections Integrated II, 2015 View details

2. Section 5.2

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Exercise 118 Page 303

Practice makes perfect

a We want to solve the given equation by factoring.

Applying the Difference of Two Squares

Did you notice that the expression on the left-hand side of the equation is a difference of two perfect squares? This can be factored using the difference of squares method.

a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.

Expression	10000x^2-64
Rewrite as Perfect Squares	(100x)^2 - 8^2
Apply the Formula	(100x+8)(100x-8)

Solving the Equation

Finally, to solve the equation we will use the Zero Product Property.

1000x^2-64=0

▼

a^2-b^2=(a+b)(a-b)

WritePow

Write as a power

(100x)^2-8^2=0

FacDiffSquares

a^2-b^2=(a+b)(a-b)

(100x+8)(100x-8)=0

ZeroProdProp

Use the Zero Product Property

lc100x+8=0 & (I) 100x-8=0 & (II)

▼

(I), (II): Solve for x

SubEqn

(I): LHS-8=RHS-8

l100x=- 8 100x-8=0

AddEqn

(II): LHS+8=RHS+8

l100x=- 8 100x=8

(I), (II):.LHS /100.=.RHS /100.

lx= -8100 x= 8100

MoveNegNumToFrac

(I): Put minus sign in front of fraction

lx=- 8100 x= 8100

(I), (II): a/b=.a /4./.b /4.

lx_1=- 225 x_2= 225

Using the Difference of Squares, we found that the solutions of the given equation are x_1=- 225 and x_2= 225.

b 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 and then simplify as much as possible.

9x^2-8=- 34x

AddEqn

LHS+34x=RHS+34x

9x^2-8+34x=0

CommutativePropAdd

Commutative Property of Addition

9x^2+34x-8=0

Now we can identify the values of a, b, and c. 9x^2+34x-8=0 ⇕ 9x^2+ 34x+( - 8)=0 We see that a= 9, b= 34, and c= - 8. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 34±sqrt(34^2-4( 9)( - 8))/2( 9)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 34±sqrt(1156-4(9)(- 8))/2(9)

Multiply

x=- 34±sqrt(1156-36(- 8))/18

MultNegNegOnePar

- a(- b)=a* b

x=- 34±sqrt(1156+228)/18

AddTerms

Add terms

x=- 34±sqrt(1444)/18

CalcRoot

Calculate root

x=- 34± 38/18

FactorOut

Factor out 2

x=2(- 17± 19)/18

ReduceFrac

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

x=- 17± 19/9

The solutions for this equation are x= - 17± 199. Let's separate them into the positive and negative cases.

x=- 17± 19/9
x_1=- 17+ 19/9	x_2=- 17- 19/9
x_1=2/9	x_2=- 36/9
x_1=2/9	x_2=- 4

Using the Quadratic Formula, we found that the solutions of the given equation are x_1= 29 and x_2=- 4.

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. 2x^2-4x+7=0 ⇕ 2x^2+( - 4)x+ 7=0 We see that a= 2, b= - 4, and c= 7. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 4)±sqrt(( - 4)^2-4( 2)( 7))/2( 2)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=4±sqrt((- 4)^2-4(2)(7))/2(2)

NegBaseToPosPow

(- a)^2 = a^2

x=4±sqrt(16-4(2)(7))/2(2)

Multiply

x=4±sqrt(16-56)/4

SubTerm

Subtract term

x=4±sqrt(- 40)/4 *

Oops! The square root of a negative number is undefined for real numbers! As we can see, using the Quadratic Formula for the given equation resulted in contradiction. Thus, there are no real solutions. For further explanation, notice that the left-hand side of the equation is a quadratic function. Let's graph it.

The solutions of the given equation are x-intercepts of this function. Notice that this parabola lies above the x-axis, thus it does not have x-intercepts. Therefore there are no real solutions of the given equation.

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 a Let's start by rearranging terms of the equation using Commutative Property of Addition. We will also multiply both sides of the equation by 5 to avoid decimals.

3.2x+0.2x^2-5=0

CommutativePropAdd

Commutative Property of Addition

0.2x^2+3.2x-5=0

MultEqn

LHS * 5=RHS* 5

x^2+16x-25=0

Now we can identify the values of a, b, and c. x^2+16x-25=0 ⇕ 1x^2+ 16x+( - 25)=0 We see that a= 1, b= 16, and c= - 25. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 16±sqrt(16^2-4( 1)( - 25))/2( 1)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 16±sqrt(256-4(1)(- 25))/2(1)

IdPropMult

Identity Property of Multiplication

x=- 16±sqrt(256-4(- 25))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 16±sqrt(256+100)/2

AddTerms

Add terms

x=- 16±sqrt(356)/2

SplitIntoFactors

Split into factors

x=- 16±sqrt(4* 89)/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=- 16± sqrt(4)* sqrt(89)/2

CalcRoot

Calculate root

x=- 16± 2 sqrt(89)/2

FactorOut

Factor out 2

x=2(- 8± sqrt(89))/2

CalcQuot

Calculate quotient

x=- 8± sqrt(89)

Using the Quadratic Formula, we found that the solutions of the given equation are x=- 8± sqrt(89). Therefore, the solutions are x_1=- 8+sqrt(89) and x_2=- 8-sqrt(89).