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
6
y = -2 ± 2√[13] Both terms in the top and 6 have a factor of 2 which can
6 be canceled out.
y = -1 ± √[13] This is our solution set in radical form.
3
{.869, -1.535} Our solution set presented as decimal approximations.
Using the Quadratic Formula will work on all types of quadratic equations, but factoring
is quicker and easier if it is a possible choice for the equation.

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