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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

2. Solving Quadratic Equations by Graphing

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Exercise 60 Page 110

Start by rewriting the equation so that all its terms are on the left-hand side. Then, simplify the equation as much as possible. Finally, recall the formula for factoring a perfect square trinomial.

-4

Practice makes perfect

We will solve the equation and then check the solutions.

Solving the Equation

Let's start by rewriting the given quadratic equation. 2x^2+16x=-32 ⇕ 2x^2+16x+32=0Let's now factor out 2 on the left-hand side, and divide both sides by 2.

2x^2+16x+32=0

SplitIntoFactors

Split into factors

2x^2+2(8)x+2(16)=0

Factor out 2

2(x^2+8x+16)=0

.LHS /2.=.RHS /2.

x^2+8x+16=0

Note that the first and last terms of the left-hand side are perfect squares, and that the middle term is twice the product of their square roots. This means that it is a perfect square trinomial and we can factor it following the corresponding formula. a^2+2ab+b^2=(a+b)^2 Let's solve the equation by factoring the perfect square trinomial on the left-hand side.

x^2+8x+16=0

Write as a power

x^2+8x+4^2=0

SplitIntoFactors

Split into factors

x^2+2(4)x+4^2=0

CommutativePropMult

Commutative Property of Multiplication

x^2+2x(4)+4^2=0

FacPosPerfectSquare

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

(x+4)^2=0

sqrt(LHS)=sqrt(RHS)

sqrt((x+4)^2)=sqrt(0)

sqrt(a^2)=± a

x+4=± 0

LHS-4=RHS-4

x=-4

Checking the Solutions

We can check our answer by substituting it for x in the given equation.

x^2+8x+16=0

x= -4

( -4)^2+8( -4)+16? =0

NegBaseToPosPow

(- a)^2 = a^2

16+8(-4)+16? =0

a(- b)=- a * b

16-32+16? =0

Add and subtract terms

0=0 ✓

Since substituting and simplifying created a true statement, we know that -4 is a solution to the equation.

Subchapter links

Exercises p.108-110