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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Study Guide and Review

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Exercise 23 Page 217

To solve the given equation, let's first isolate x^2.

8x^2+16=0

LHS-16=RHS-16

8x^2=- 16

.LHS /8.=.RHS /8.

x^2=- 2

We cannot find any real square roots of a negative number, so we will need to use the fact that sqrt(- a)= isqrt(a) to solve for x.

x^2=- 2

sqrt(LHS)=sqrt(RHS)

x=± sqrt(- 2)

sqrt(- a)= isqrt(a)

x=± i sqrt(2)

CommutativePropAdd

Commutative Property of Addition

x=± sqrt(2)i

Subchapter links

1 Solving Quadratic Equations by Factoring p.217

2 Complex Numbers p.217

3 The Quadratic Formula and the Discriminant p.218

4 Transformations of Quadratic Graphs p.218

5 Quadratic Inequalities p.219

Practice Test p.219