Solutions
Name __________________________
Block ____ Date ____________
Algebra: 10.2.4: How Many Solutions?
Bell Work: Simplify and then solve each equation.
100
1
6 1
a.
2
𝑥+
3
=5
b.
0.03𝑥 + 0.05 = −0.04
3𝑥 + 5 = −4
3𝑥 + 2 = 30
3𝑥 = −9
3𝑥 = 28
𝒙 = −𝟑
𝟐𝟐
𝒙=
𝟑
10
c.
0.1𝑥 2 − 0.3𝑥 − 2.8 = 0
𝑥 2 − 3𝑥 − 28 = 0
(𝑥 − 7)(𝑥 + 4) = 0
𝒙 = 𝟕 and 𝒙 = −𝟒
−7 −7𝑥 −28
𝑥 𝑥 2 +4𝑥
𝑥
10-60. THE NUMBER OF SOLUTIONS
With your partner, solve the equations below and note how many solutions there are. Express your
answers in both exact form and approximate decimal form. Look for patterns among those with no
solution and those with only one solution.
(x + 4)2 = 20
a.
𝑥 + 4 = ±√20
2 Solutions 𝑥 + 4 = ±� 4 ∙ 5
(7x − 5)2 = −2
b.
𝑥 + 4 = ±2√5
𝒙 = −𝟒 ± 𝟐√𝟓
𝒙 ≈ 𝟎. 𝟒𝟒 𝒂𝒂𝒂 𝒙 ≈ −𝟖. 𝟒𝟒
d.
(5 − 10x)2 = 0
e.
(x + 2)2 = −10
𝑥 + 2 = ±√−10
7𝑥 − 5 = ±√−2
No Real Solutions
(2x − 3)2 = 49
c.
2 Solutions
2𝑥 − 3 = ±7
2𝑥 = 3 ± 7
3±7
𝑥=
2
5 − 10𝑥 = 0
−10𝑥 = −5
𝟏
𝒙=
1 Solution
𝟐
No Real Solutions
𝑥=
𝑥=
3+7
=𝟓
2
3−7
= −𝟐
2
f.
(x + 11)2 + 5 = 5
(𝑥 + 11)2 = 0
𝑥 + 11 = 0
𝒙 = −𝟏𝟏
1 Solution
10-61. Use the patterns you found in problem 10-60 to determine quickly how many solutions there are
for each quadratic equation below. You do not need to solve the equations.
a.
(5m − 2)2 + 6 = 0
No Real Solutions
b.
(4 + 2n)2 = 0
1 Solution
c.
11 = (7 + 2x)2
2 Solutions
+4
10-62. How can you tell how many solutions an equation in standard form, ax2 +bx + c = 0
has? Explore that question by completing this investigation:
a.
1
Solve the quadratic equation 2x2 − 3x + 2 = 0. Write the solution in both exact form and approximate
decimal form.
b.
1
Change the 2 at the end of the equation to another number so that you have an equation with no real
solutions.
What would the
c.
1
2
have to be so there is only one solution?
Explain how you can tell from the standard form, ax2 + bx + c = 0, whether the equation will have
zero, one, or two real solutions?
10-63. How many solutions do the equations below have? How can you represent the solutions?
a.
9x2 + 4x − 6 = y
b.
� 𝑥 + 1� = 𝑦
3
8
2
1
10-64. Use the graph and table in the calculator to find solutions to 2x2 − 3x + = 0
2
10-65. Consider the equation |2𝑥 − 5| = 9 .
a.
With your team, solve |2𝑥 − 5| = 9. Record your work
carefully as you go. Check your solution(s).
b.
Can an absolute value equation have no solution? With your
partner, create an absolute value equation that has no
solution. How can you be sure there is no solution?
c.
Likewise, create an equation with an absolute value that will
have only one solution. Justify why it will have only one
solution.
2𝑥 − 5 = 9
2𝑥 = 14
𝒙=𝟕
2𝑥 − 5 = −9
2𝑥 = −4
𝒙 = −𝟐
Yes.
|2𝑥 + 3| = −6
Absolute Value cannot
equal a negative number
|2𝑥 + 3| = 0
Zero has no sign so ±0
produces only 1 equation
10-67. Solve these equations, if possible. Each time, be sure you have found all possible solutions by
graphing in the calculator. Show sufficient work for full credit.
a.
(x + 4)2 = 49
b.
3√𝑥 + 2 = 12
10-68. Is x = −4 a solution to
c.
d.
1
3
2
𝑥
+
3
10
=
13
10
5(2x − 1) − 2 = 13
(2𝑥 + 5) > −1 ? Explain how you know.
1
[2(−4) + 5] = −𝟏
3
10-69. Factor each of the following expressions completely. Be sure to look for any common factors.
a.
4x2 − 12x
b.
3y2 + 6y + 3
Use Generic
Rectangles on b, c
and d for full credit
c.
2m3 + 7m2 + 3m
d.
3x2 + 4x − 4
10-70. Write and solve an equation to answer the question below. Remember to define any variables you use.
Pierre's Ice Cream Shoppe charges $1.19 for a scoop of ice cream and $0.49 for each topping.
Gordon paid $4.55 for a three-scoop sundae. How many toppings did he get?
t = number of toppings, 1.19(3) + 0.49t = 4.55, and t = 2
10-71. Examine the tile pattern below. Based on the information provided for Figures 1 through 4,
answer the questions below.
a.
Represent the number of tiles with a
table and a rule.
b.
Find the number of tiles in Figure
5. Explain how you found your
answer.
10-72. Solve: 2x2 − 19x + 9 = 0 twice, using Factoring and the Quadratic Formula.
Verify that the solutions from both methods are the same.
b. Quadratic Formula 𝑥 =
a. Factoring:
2x2 − 19x + 9 = 0
−1 −1𝑥
2x2 − 19x + 9 = 0
9
2𝑥 2𝑥 2 −18𝑥
2𝑥 − 1 = 0
2𝑥 = 1
𝟏
𝒙=
𝟐
𝑥
𝑥=
−9
𝑥−9=0
𝒙=𝟗
−𝑏±√𝑏2 −4𝑎𝑎
2𝑎
19 ± �(−19)2 − 4(2)(9)
2(2)
𝑥=
19 ± √361 − 72
4
19 ± √289
𝑥=
4
𝑥=
19 ± 17
4
𝑥=
𝑥=
19 + 17
=𝟗
4
19 − 17 𝟏
=
4
𝟐
10-73. Write an inequality to represent this situation.
Adult tickets to the game are sold for $7 and student tickets are sold for $5. What
combination of tickets must be sold to have total sales of at least $5000?