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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

8. Graphing Radical Functions

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Exercise 72 Page 420

Make sure you write all the terms on the left-hand side of the equation and simplify as much as possible before using the Quadratic Formula.

-3± 3sqrt(5)/2

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

Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible.

3x^2+9x=27

LHS-27=RHS-27

3x^2+9x-27=0

Now, we can identify the values of a, b, and c. 3x^2+9x-27=0 ⇕ 3x^2+( 9)x+( - 27)=0 We see that a= 3, b= 9, and c= - 27. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( 9)±sqrt(( 9)^2-4( 3)( - 27))/2( 3)

▼

Solve for x and Simplify

Multiply

x=-9±sqrt((9)^2-(12)(- 27))/6

Calculate power

x=-9±sqrt(81-(12)(- 27))/6

MultNegNegOnePar

- a(- b)=a* b

x=-9±sqrt(81+324)/6

Add terms

x=-9±sqrt(405)/6

SplitIntoFactors

Split into factors

x=-9±sqrt(81* 5)/6

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

x=-9± sqrt(81)* sqrt(5)/6

Calculate root

x=-9± 9sqrt(5)/6

Factor out 2

x=3(-3± 3sqrt(5))/3*2

CancelCommonFac

Cancel out common factors

x=-3± 3sqrt(5)/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= -3± 3sqrt(5)2. Therefore, the solutions are x_1= -3+3sqrt(5)2 and x_2= -3-3sqrt(5)2.

Subchapter links

Lesson Check p.418

Practice and Problem-Solving Exercises p.418-420

Standardized Test Prep p.420

Mixed Review p.420