GPS Algebra SupportName
Unit 5B Review – Solving Quadratics Guided Notes

Factoring trinomials where a > 1
I TRY / YOU TRY
Problem / 3x2 – 13x - 30 / 2x2 + 5x – 12
Step 1: Separate the middle by finding two numbers that equal the product of a and c and add to your b. / a • c = 3 • -30 = -90
b = -13
90
1 90
2 45
3 30
5 18
6 15
9 10
-18 and 5
3x2 – 18x + 5x – 30 / a • c = ____• -12 = -24
b = ____
24
1 ___
2 12
3 8
___ 6
-3 and ___
2x2 – 3x + ___ – 12
Step 2: Group first two terms and second two terms. / (3x2 – 18x) + (5x – 30) / (2x2 - ___ ) + (___ - 12)
Step 3: Find the largest number and variable that divides into each group (the GCF) and take it out. Your remaining parentheses should match. / 3x(x – 6) + 5(x – 6) / x(2x – 3) + ___(2x – 3)
Step 4: Bring down one set of the matching parentheses and group your GCFs in the other parentheses. / (3x + 5)(x – 6) / (x + ___)(2x – 3)
Step 5: Circle your answer! Factoring is just finding the expressions that multiply together; it is not solving. / (3x + 5)(x – 6)

Factor the following using the method above:

5x2 – 24x – 5 / 2x2 – 31x + 18
Step 1 / a • c = ___ • ___ = -25
b = ___
25
1 ___
5 ___
___ and ___
5x2 ______- 5
Step 2 / (5x 2 ______) + (______- 5)
Step 3 / ___(______) ___(______)
Step 4 / (____ + ____)(____ - ____)
Step 5
Solving Quadratics by Factoring – use when you have an x2 and an x
I TRY / YOU TRY
Problem / 2x2 – 6x + 17 = 7x + 2 / 3x2 + 9x + 2 = 2x
Step 1: Set equation equal to zero. Do this by adding or subtracting everything on the side of the equation opposite the x2. Remember to do the same thing to both sides. Combine like terms. If there are no like terms, just write the terms next to each other in the correct order. / 2x2 – 6x + 17 = 7x + 2
-7x - 2 -7x - 2
2x2 – 13x + 15 = 0 / 3x2 + 9x + 2 = 2x
- ___ - ___
3x2 + 7x + 2 = 0
Step 2: Factor your equation. Always factor out any common factors first. If the equation still begins with something other than 1, use the steps above. / 2x2 – 13x + 15
a • c = 2 • 15 = 30, b = -13
-10 and -3
2x2 – 10x – 3x + 15
(2x2 – 10x) + (-3x + 15)
2x(x – 5) + -3(x – 5)
(2x – 3)(x – 5) = 0 / 3x2 + 7x + 2
a • c = 3 • 2 = 6, b = 7
___ and ___
3x2 + ___ + ___ + 2
(3x2 + ___) + (___ + 2)
___(___ + ___) + ___(___ + ___)
(___ + ___)(___ + ___)
Step 3: Set each factor equal to 0. Your factors are the expressions in parentheses. Solve each equation for x. / 2x – 3 = 0 x – 5 = 0
+ 3 + 3 + 5 + 5
2x = 3 x = 5
2 2
x = 3/2 / 3x + 1 = 0 x + 2 = 0
- - - - _
3x = ___ x = ___
3 3
x = ___
Step 4: Circle your 2 x = … ‘s . These are your answers. Solving is finding a value for x. /
x = 3/2 x = 5 / x = ___ x = ___

Solve the following using the methods above

6x2 = 2x / 2x2 + 4x – 2 = 2 – x2
Step 1 / 6x2 = 2x
- -___
Step 2
Step 3 / ___ = 0 ___ - ___ = 0
Step 4
Solving Quadratics by Square-Roots – use when you have an x2 or an (x±k)2 but not both x2 and x
I TRY / YOU TRY
Problem / 2(x + 2)2 + 3 = 53 / 3(x – 4)2 – 1 = 59
Step 1: Get the squared expression by itself. Add and subtract anything outside the parentheses and then divide or multiply by anything that is in front of the expression. / 2(x + 2)2 + 3 = 53
- 3 -3
2(x + 2)2 = 50
2 2
(x + 2)2 = 25 / 3(x – 4)2 – 1 = 59
+ +
3(x – 4)2 = 60
______
(x – 4)2 = 20
Step 2: Square-root each side. If the number is a perfect square, make sure you take the + and – of the answer. If the number is not a perfect square, simplify the radical and put a ± in front. / (x + 2)2 = 25

x + 2 = 5, -5 / (x – 4)2 = 20

x – 4 = ± ___
Step 3: If x is not by itself, add or subtract the number with it to get it by itself. Remember that if you had whole number answers, you can combine like terms. If you had a radical, just put the added number in front of the ±. / x + 2 = 5 x + 2 = -5
- 2 -2 - 2 -2
x = 3 x = - 7 / x – 4 = ± ___
+ 4 + 4 __
x = 4 ± ___
Step 4: Circle your answer. Solving is finding a value for x. / x = 3 x = -7 / x = ______

Solve the following using the methods above

5x2 – 7 = 73 / 2x2 + 4x – 2 = 2 – x2
Step 1 / 5x2 – 7 = 73
+ +___
Step 2
Step 3
Step 4
Solving Quadratics by Quadratic Formula – use when you have an x2 and an x
(same times when you might use factoring)
I TRY / YOU TRY
Problem / 2x2 – 6x + 17 = 7x + 2 / 3x2 + 9x + 2 = 2x
Step 1: Set equation equal to zero. Do this by adding or subtracting everything on the side of the equation opposite the x2. Remember to do the same thing to both sides. / 2x2 – 6x + 17 = 7x + 2
-7x - 2 -7x - 2
2x2 – 13x + 15 = 0 / 3x2 + 9x + 2 = 2x
- ___ - ___
3x2 + 7x + 2 = 0
Step 2: Label a, b, and c. Remember a is the number in front of x2, b is the number of x’s (not x2s), and c is the number by itself. / 2x2 – 13x + 15 = 0
a = 2, b = -13, c = 15 / 3x2 + 7x + 2 = 0
a = ___, b = ___, c = ___
Step 3: Plug a, b, and c into the correct places in the formula:
/ /
Step 4: Simplify the radical. Combine like terms, if possible. If not possible, simplify the fraction. / =
=
Step 5: Circle your answers. / x = 5 x = 3/2

Solve the following using the methods above

5x2 – 7x = 12 / 3x2 = 21
Step 1 / 5x2 – 7x = 12
- -___
Step 2 / a = ____, b = ____, c = ____
Step 3 / /
Step 4
Step 5

Solve the following using any method:

  1. 3(x – 1)2 = 184. 3x2 + 4 = 2x – 7
  1. 2x2 + x = 285. 7x2 – 12x + 14 = x2 + 8x
  1. 5x2 = 4x6. 2(x – 7)2 – 4 = 14

Solving Quadratic Inequalities
I TRY / YOU TRY
Problem / x2 – 7x ≤-10 / 4x2 > 12x
Step 1: Make inequality an equation. Solve quadratic using all techniques above. / x2 – 7x = -10
+ 10 +10
x2 – 7x + 10 = 0
(x – 2)(x – 5) = 0
x – 2 = 0 x – 5 = 0
+ 2 + 2 + 5 + 5
x = 2 x = 5 / 4x2 = 12x
- - _
= 0
___(___ - ___) = 0
___ = 0 ___ - ___ = 0
x = ___ x = ___
Step 2: Use solutions as endpoints. Plot on number line using solid dots if ≤ or ≥ or hollow dots if < or >. / 0 2 4 6 8 / -4 -2 0 2 4
Step 3: Pick a value from each section of your number line.
1 to the left of both points
1 from between the 2 points
1 to the right of the both points
Test these points in the original inequality to see if they make it true. / x2 – 7x ≤ -10
0, 4, 6
(0)2 – 7(0) ≤ 10
0 ≤ -10 X
(4)2 – 7(4) ≤ -10
-12 ≤ -10 √
(6)2 – 7(6) ≤ -10
-6 ≤ -10 X
Step 4: Shade the number line in the sections that were true. / 0 2 4 6 8

/ -4 -2 0 2 4
Step 5: Write inequalities that relate to the shaded region of your number line. Remember it is an “or” situation if you shade the ends and an “and” situation if you shade the middle. / x ≥ 2 and x ≤ 5
2 ≤ x ≤ 5

Solve the following inequalities

  1. 2x2 + x < 283. 7x2 – 12x + 14 ≥ x2 + 8x
  1. 5x2 > 4x4. 2(x – 7)2 – 4 ≤ 14