Exercise 63 Page 139

We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so only terms with x are on one the left of the equation. 6x^2 - 17x + 12 = 0 ⇕ 6x^2 - 17x = - 12 Now, let's divide each side by 6 so the coefficient of x^2 will be 1.

6x^2 - 17x = - 12

DivEqn

.LHS /6.=.RHS /6.

6x^2 - 17x/6=- 12/6

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Simplify left-hand side

WriteDiffFrac

Write as a difference of fractions

6x^2/6-17x/6=- 12/6

MovePartNumRight

a* b/c=a/c* b

6/6x^2-17/6x=- 12/6

CalcQuot

Calculate quotient

x^2-17/6x=- 2

In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=- 176. Let's now calculate ( b2 )^2.

( b/2 )^2

Substitute

b= - 17/6

( - 17/6/2 )^2

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Simplify

MoveNegNumToFrac

Put minus sign in front of fraction

(- 17/6/2 )^2

NegBaseToPosBase

(- a)^2=a^2

(17/6/2 )^2

Rewrite

Rewrite 17/6/2 as 17/6÷ 2

( 17/6 ÷ 2 )^2

WriteAsFrac

Write as a fraction

( 17/6 ÷ 2/1 )^2

DivFracByFracDiv

a/b÷c/d=a/b*d/c

( 17/6 * 1/2 )^2

MultFrac

Multiply fractions

( 17/12 )^2

PowQuot

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

289/144

Next, we will add ( b2 )^2= 289144 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.

x^2-17/6x=- 2

AddEqn

LHS+289/144=RHS+289/144

x^2-17/6x+ 289/144=- 2+ 289/144

FacNegPerfectSquare

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

(x-17/12)^2=- 2+289/144

WriteAsFrac

Write as a fraction

(x-17/12)^2=- 288/144+289/144

AddFrac

Add fractions

(x-17/12)^2=1/144

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-17/12)^2)=sqrt(1/144)

CalcRoot

Calculate root

x-17/12=± 1/12

AddEqn

LHS+17/12=RHS+17/12

x=17/12± 1/12

The solutions for this equation are x= 1712± 112. Let's separate them into the positive and negative cases.

x=17/12± 1/12
x_1=17/12 + 1/12	x_2=17/12 - 1/12
x_1=18/12	x_2=16/12
x_1=3/2	x_2=4/3

We found that the solutions of the given equation are x_1= 32 and x_2= 43.