Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 10.2
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Exercise 72 Page 494

The expression on the left-side of the equation is given in the standard form of a quadratic function. Start by identifying the values of a, b, and c.

Solutions: x=- 1/2, x=- 9
Are the solutions the same? Yes.

Practice makes perfect

We want to solve the given equation for x twice — once using the Zero Product Property and once using the Quadratic Formula. Then we will compare the answers.

Using Zero Product Property

We want to solve the given equation using the Zero Product Property. To do this, we need to rewrite the given equation in the factored form. We can start by using factoring.

Factoring

Here we have a quadratic equation of the form ax^2+bx+c=0, where |a| ≠ 1 and there are no common factors. To factor this equation we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 2x^2+19x+9=0 We have that a= 2, b=19, and c=9. There are now three steps we need to follow in order to rewrite the above equation.
  1. Find a c. Since we have that a= 2 and c=9, the value of a c is 2* 9=18.
  2. Find factors of a c. Since a c=18, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=19, which is positive, those factors will need to be positive so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 & 18 & 1 + 18 &19 2 &9 &2 + 9 &11 3 &6 &3 + 6 &9

  1. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 2x^2+19x+9=0 ⇕ 2x^2 + 1x + 18x+9=0
Finally, we will factor the last equation obtained.
2x^2+1x+18x+9=0
x(2x+1)+18x+9=0
x(2x+1)+9(2x+1)=0
(2x+1)(x+9)=0
Now the equation is written in a factored form.

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
(2x+1)(x+9)=0
lc2x+1=0 & (I) x+9=0 & (II)
â–Ľ
(I): Solve for x
l2x=- 1 x+9=0
lx=- 12 x+9=0
lx_1=- 12 x_2=- 9
We found that x=- 12 or x=- 9.

Using Quadratic Formula

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 Since a= 2, b= 19, and c= 9, we will substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 19±sqrt(19^2-4( 2)( 9))/2( 2)
â–Ľ
Solve for x and Simplify
x=- 19±sqrt(361-4(2)(9))/2(2)
x=- 19±sqrt(361-72)/4
x=- 19±sqrt(289)/4
x=- 19± 17/4
The solutions for this equation are x= - 19± 174. Let's separate them into the positive and negative cases.
x=- 19± 17/4
x_1=- 19+17/4 x_2=- 19-17/4
x_1=- 2/4 x_2=- 36/4
x_1=- 1/2 x_2=- 9

Using the Quadratic Formula, we found that the solutions are x_1=- 12 and x_2=- 9. Notice that those are the same solutions as obtained previously by factoring and then using Zero Product Property. Both methods are equally good, so they give the same solutions.