INSTRUCTOR GUIDANCE EXAMPLE: Week Four Discussion

Solving Quadratic Equations

#79 pg 637: Solve by Factoring

There is another way this problem could be solved (by completing the square which has

already been done to it) but our instructions say to solve by factoring. This will require

us to multiply out the left side and then subtract 9/4 from both sides to leave the right side

zero.

(p + ½ )2 = 9/4 First we need to expand the left side by FOIL.

p2 + p + ¼ = 9/4 Subtract 9/4 from both sides.

p2 + p – 2 = 0 Since ¼ – 9/4 = -8/4 = -2 we now are free of the fractions.

(p + 2)(p – 1) = 0 Left side is factored.

p + 2 = 0 or p – 1 = 0 Using the Zero Factor Property.

p = -2 or p = 1 Our solutions.

{-2, 1} Solution set presented.

Check: (p + ½ )2 = 9/4 (p + ½ )2 = 9/4

(-2 + ½ )2 = 9/4 (1 + ½)2 = 9/4

(-3/2)2 = 9/4 (3/2)2 = 9/4

9/4 = 9/4 9/4 = 9/4

#87 pg 637

-x2 + x + 6 = 0 Factor -1 out of all terms first.

-1(x2 – x – 6) = 0 Divide both sides by -1

x2 – x – 6 = 0 Ready for factoring.

(x – 3)(x + 2)= 0 Left side is factored.

x – 3 = 0 or x + 2 = 0 Using the Zero Factor Property.

x = 3 or x = -2 Our solutions.

{-2, 3} Solution set presented.

Check: -x2 + x + 6 = 0 -x2 + x + 6 = 0

-(-2)2 + (-2) + 6 = 0 -(3)2 + 3 + 6 = 0

-4 – 2 + 6 = 0 -9 + 3 + 6 = 0

-6 + 6 = 0 -6 + 6 = 0

0 = 0 0 = 0

#47 pg 646: Solve using Quadratic Formula

3y2 + 2y – 4 = 0 a = 3, b = 2, c = -4 Discriminant is b2 – 4ac

which is 22 – 4(3)(-4) = 52 so we have two real solutions.

y = -(2) ± √[22 – 4(3)(-4)] All values put into the formula in parenthesis.

2(3)

y = -2 ± √[4 + 48] Simplification begins.

6

y = -2 ± √[52] Need to simplify the radical next: 52 = 4⋅13

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