Page 672 - Algebra 1

Example
3
Find the roots. 20-2x2 =70-20x SOLUTION
2x2 - 20x + 50 = 0 2(x2 - 10x + 25) = 0 2(x - 5)(x - 5) = 0
Set the equation equal to zero. Factor out the GCF.
Factor the trinomial expression.
Finding the Roots by Factoring Out the GCF
Disregard the factor of 2, since it can never equal 0.
The factor (x - 5) appears twice, but it only needs to be set to equal zero once.
x-5=0 x=5
Check
20-2x2 = 70-20x 20 - 2(5)2 70 - 20(5)
20-2(25) 70-100 20-50 -30
-30 = -30 ✓ The root is 5.
Finding the values of x that satisfy the quadratic equation is another way of finding the roots of a quadratic equation.
Application: Gardening
The area of a rectangular garden is 51 square yards. The length is 14 yards more than the width. What are the length and width of the garden?
SOLUTION
Let w be the width and w + 14 be the length.
Caution
When checking your answers, use the
original equation, not the one that has been rearranged. Also, use the order of operations when simplifying each side of the equation.
Example
4
A = lw
51 = (w + 14)w 51 = w2 + 14w
0 = w2 + 14w - 51
0 = (w + 17)(w - 3) w + 17 = 0 w - 3 = 0 w = -17 w = 3
Area formula
Substitute known values into the equation. Distribute.
Write the equation into standard form. Factor.
Use the Zero Product Property. Solve.
Math Reasoning
Justify How could you check your answers?
Because the width must be a positive number, the only possible solution is 3 yards. Since the width is 3 yards, the length is w + 14. So the length is
3 + 14, which is 17 yards.
Lesson 98 657