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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 73 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.

-1±sqrt(61)/10

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.

5x^2+x=3

LHS-3=RHS-3

5x^2+x-3=0

Now, we can identify the values of a, b, and c. 5x^2+x-3=0 ⇕ 5x^2+( 1)x+( - 3)=0 We see that a= 5, b= 1, and c= - 3. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( 1)±sqrt(( 1)^2-4( 5)( - 3))/2( 5)

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Solve for x and Simplify

Multiply

x=-1±sqrt((1)^2-20(- 3))/10

Calculate power

x=-1±sqrt(1-20(- 3))/10

MultNegNegOnePar

- a(- b)=a* b

x=-1±sqrt(1+60)/10

Add terms

x=-1±sqrt(61)/10

Using the Quadratic Formula, we found that the solutions of the given equation are x= -1± sqrt(61)10. Therefore, the solutions are x_1= -1+sqrt(61)10 and x_2= -1-sqrt(61)10.

Subchapter links

Lesson Check p.418

Practice and Problem-Solving Exercises p.418-420

Standardized Test Prep p.420

Mixed Review p.420