Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. Quadratic Equations
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Exercise 3 Page 229

Make sure you rewrite the equation leaving all the terms on one side and that you factor out the greatest common factor if it exists.

- 23, 1

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by factoring the left-hand side of the equation.
3x^2-x-2=0
3x^2-3x+2x-2=0
â–Ľ
Factor out 3x & 2
3x(x-1)+2x-2=0
3x(x-1)+2(x-1)=0
(3x+2)(x-1)=0

Solving

To solve this equation, we will apply the Zero Product Property.
(3x+2)(x-1)=0
lc3x+2=0 & (I) x-1=0 & (II)
l3x=- 2 x-1=0
lx=- 23 x-1=0
lx_1=- 23 x_2=1
We found that the solutions to the equation are - 23 and 1.

Checking Our Answer

Checking our answer
We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=- 23.
3x^2-x-2=0
3( -2/3) ^2 -( -2/3)-2? =0
â–Ľ
Evaluate left-hand side
3(4/9)-(-2/3) -2? =0
3* 4/9-(-2/3)-2? =0
4/3-(-2/3)-2? =0
4/3+2/3-2? =0
6/3-2? =0
2-2? =0
0=0 âś“
Substituting and simplifying created a true statement, so we know that x=- 23 is a solution of the equation. Let's move on to x=1.
3x^2-x-2=0
3( 1) ^2 - 1-2? =0
â–Ľ
Evaluate left-hand side
3(1)-1-2? =0
3-1-2? =0
0=0 âś“
Again, we created a true statement. x=1 is indeed a solution of the equation.