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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Solving Quadratic Equations by Using the Quadratic Formula
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Exercise 64 Page 139

To complete the square, make sure all the variable terms are on one side of the equation. Then, divide both sides of the equation by a so the coefficient of x^2 is 1.

7.4, 1.6

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 - 9x = - 12 In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=- 9. Let's now calculate ( b2 )^2.

( b/2 )^2
Substitute

b= - 9

( - 9/2 )^2
â–¼
Simplify
MoveNegNumToFrac

Put minus sign in front of fraction

(- 9/2)^2
NegBaseToPosBase

(- a)^2=a^2

(9/2)^2
PowQuot

(a/b)^m=a^m/b^m

9^2/2^2
CalcPow

Calculate power

81/4

Next, we will add ( b2 )^2= 814 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.

x^2 - 9x = - 12
AddEqn

LHS+81/4=RHS+81/4

x^2- 9x + 81/4=- 12+ 81/4
FacNegPerfectSquare

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

(x-9/2)^2=- 12 + 81/4
WriteAsFrac

Write as a fraction

(x-9/2)^2=- 48/4 + 81/4
AddFrac

Add fractions

(x-9/2)^2=33/4
SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-9/2)^2)=sqrt(33/4)
CalcRoot

Calculate root

x-9/2=± sqrt(33/4)
RootQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

x-9/2=± sqrt(33)/2
AddEqn

LHS+9/2=RHS+9/2

x=9/2± sqrt(33)/2

The solutions for this equation are x= 92± sqrt(33)2. Let's separate them into the positive and negative cases.

x=9/2± sqrt(33)/2
x_1=9/2 + sqrt(33)/2 x_2=9/2 - sqrt(33)/2
x_1=9 + sqrt(33)/2 x_2=9 - sqrt(33)/2
x_1≈ 7.4 x_2≈ 1.6

We found that the solutions of the given equation are x_1≈ 7.4 and x_2≈ 1.6.

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arrow_right Exercises p.137-139
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