McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
Standardized Test Practice

Exercise 6 Page 914

Use the given roots to write the equation in factored form. Then multiply and simplify to obtain the expression in standard form.

J

Practice makes perfect

We want to determine which of the given quadratic equations has the roots 12 and 13. To do so, we can write a quadratic equation in factored form using the given roots. Then we will rewrite it to be in standard form by multiplying the factors. rr Factored Form: & a(x- p)(x- q)=0 Standard Form: & ax^2+ bx+ c=0In the factored form, p and q are the roots of the equation. Since we are told that the roots are 12 and 13, we can partially write the factored form of our equation. a( x- 12 ) ( x- 13 )=0 Since a does not have any effect on the roots, we can choose any value. For simplicity, and in order to have integer coefficients, we will let a= 6. This is a common multiple of both denominators of the given roots and will allow us to eliminate the fractions when we distribute. Let's substitute this value and use the Distributive Property to obtain the standard form.

a( x- 1/2 ) ( x-1/3 )=0
6( x- 1/2 ) ( x-1/3 )=0
â–¼
Distribute 6
2(3)( x- 1/2 ) ( x-1/3 )=0
2( x- 1/2 ) (3)( x-1/3 )=0
( 2x- 1 ) (3)( x-1/3 )=0
(2x-1)(3x-1)=0
â–¼
Multiply parentheses
2x(3x-1)-1(3x-1)=0
6x^2-2x-1(3x-1)=0
6x^2-2x-3x+1=0
6x^2-5x+1=0

Note that we obtained the equation from option J. Therefore, J is the correct answer.

Alternative Solution

Check the Answers by Substituting
We could also check which of the given quadratic equations has the roots 12 and 13 by substituting the roots into the equations. If we obtain true statements after substituting both of the values into the equation, 12 and 13 are the roots of the equation. Otherwise, these values are not the roots of the equation. Let's do it!

Equation Substitute Evaluate
5x^2-5x-2=0 5( 1/2)^2-5( 1/2)-2? =0 - 13/4 ≠ 0
5( 1/3)^2-5( 1/3)-2? =0 - 28/9≠ 0
5x^2-5x+1=0 5( 1/2)^2-5( 1/2)+1? =0 - 1/4≠ 0
5( 1/3)^2-5( 1/3)+1? =0 - 1/9≠ 0
6x^2+5x-1=0 6( 1/2)^2+5( 1/2)-1? =0 3≠ 0
6( 1/3)^2+5( 1/3)-1? =0 4/3≠ 0
6x^2-5x+1=0 6( 1/2)^2-5( 1/2)+1? =0 0=0
6( 1/3)^2-5( 1/3)+1? =0 0=0

We obtained true statements only for the equation 6x^2-5x+1=0. Therefore, this is the only equation that has the roots 12 and 13 and our answer is correct.