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McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Solving Quadratic Equations by Using the Quadratic Formula
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Exercise 18 Page 137

Identify a, b and c.

1.9, 0.5

Practice makes perfect

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. 6x^2-12x+1=0 ⇕ 6x^2 + ( - 12)x + 1=0 We see that a= 6, b= - 12, and c= 1. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a
SubstituteValues

Substitute values

x=- ( - 12)±sqrt(( - 12)^2-4( 6)( 1))/2( 6)
â–¼
Solve for x and Simplify
NegNeg

- (- a)=a

x=12±sqrt((- 12)^2-4(6)(1))/2(6)
CalcPow

Calculate power

x=12±sqrt(144-4(6)(1))/2(6)
Multiply

Multiply

x=12±sqrt(144-24)/12
SubTerm

Subtract term

x=12±sqrt(120)/12
SplitIntoFactors

Split into factors

x=12±sqrt(4 * 30)/12
SqrtProd

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

x=12± sqrt(4)* sqrt(30)/12
CalcRoot

Calculate root

x=12± 2 sqrt(30)/12
FactorOut

Factor out 2

x=2(6± sqrt(30))/12
CancelCommonFac

Cancel out common factors

x=6± sqrt(30)/6

The solutions for this equation are x= 6± sqrt(30)6. Let's separate them into the positive and negative cases.

x=6 ± sqrt(30)/6
x_1=6+sqrt(30)/6 x_2=6 - sqrt(30)/6
x_1≈ 1.9 x_2≈ 0.5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1 ≈ 1.9 and x_2≈ 0.5.

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