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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

4. Arithmetic Series

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Exercise 61 Page 593

To complete the square, make sure all the variable terms are on one side of the equation.

x=-5 ± i sqrt(10)

Practice makes perfect

We want to solve the quadratic equation by completing the square. Note that all terms with x are on one side of the equation. x^2+10x+40=5In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=10. Let's now calculate ( b2 )^2.

( b/2 )^2

b= 10

( 10/2 )^2

▼

Evaluate

Calculate quotient

5^2

Calculate power

25

Next, we will add ( b2 )^2=25 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.

x^2+10x + 40=5

LHS-15=RHS-15

x^2+10x+25=-10

SplitIntoFactors

Split into factors

x^2+2(5)x+25=-10

CommutativePropMult

Commutative Property of Multiplication

x^2+2x(5)+25=-10

Write as a power

x^2+2x(5)+5^2=-10

FacPosPerfectSquare

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

(x+5)^2=-10

sqrt(LHS)=sqrt(RHS)

sqrt((x+5)^2)=sqrt(-10)

SplitIntoFactors

Split into factors

sqrt((x+5)^2)=sqrt(-1 (10))

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

sqrt((x+5)^2)=sqrt(-1) * sqrt(10)

sqrt(a^2)=± a

x+5=± sqrt(-1) * sqrt(10)

sqrt(- 1)= i

x+5=± i sqrt(10)

LHS-5=RHS-5

x=-5 ± i sqrt(10)

Both x=-5 + i sqrt(10) and x=-5 - i sqrt(10) are solutions of the equation.

Subchapter links

Lesson Check p.591

Practice and Problem-Solving Exercises p.591-593

Standardized Test Prep p.593

Mixed Review p.593