McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. The Quadratic Formula and the Discriminant
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Exercise 5 Page 194

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.

(1.5, - 0.2)

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.

10x^2-3=13x
10x^2-13x-3=0

Now, we can identify the values of a, b, and c. 10x^2-13x-3=0 ⇕ 10x^2+( - 13)x+( - 3)=0 We see that a= 10, b= - 13, and c= - 3. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a
x=- ( - 13)±sqrt(( - 13)^2-4( 10)( - 3))/2( 10)
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Solve for x and Simplify
x=13±sqrt((- 13)^2-4(10)(- 3))/2(10)
x=13±sqrt(169-4(10)(- 3))/2(10)
x=13±sqrt(169-40(- 3))/20
x=13±sqrt(169+120)/20
x=13±sqrt(289)/20
x=13± 17/20

The solutions for this equation are x= 13± 1720. Let's separate them into the positive and negative cases.

x=13± 17/20
x_1=13+17/20 x_2=13-17/20
x_1=30/20 x_2=- 4/20
x_1=1.5 x_2=- 0.2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=1.5 and x_2=- 0.2.