Skip to main content

Precalculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

TBIL Community

🔗

Section 1.5 Quadratic Equations (EQ5)

🔗

Subsection 1.5.1 Activities

🔗

Definition 1.5.1.

🔗
🔗
A quadratic equation is of the form:
ax2+bx+c=0
🔗
where a and b are coefficients (and a ≠0), x is the variable, and c is the constant term.
🔗

Activity 1.5.2.

🔗
🔗
Which of the following is a quadratic equation?
  1. 6−x2=3x
  2. (2x−1)(x+3)=0
  3. 4(x−3)+7=0
  4. (x−4)2+1=0
  5. 5x2−3x=17−4x
Answer.
A, B, D, and E
🔗

Definition 1.5.3.

🔗
To solve a quadratic equation, we will need to apply the zero product property, which states that if a⋅b=0, then either a=0 or b=0. In other words, you can only have a product of 0 if one (or both!) of the factors is 0.
🔗

Activity 1.5.4.

🔗
In this activity, we will look at how to apply the zero product property when solving quadratic equations.
🔗
(a)
🔗
🔗
Which of the following equations can you apply Definition 1.5.3 as your first step in solving?
  1. 2x2−3x+1=0
  2. (2x+1)(x+1)=0
  3. 3x2−4=6x
  4. x(3x+5)=0
Answer.
B and D
🔗
(b)
🔗
🔗
Suppose you are given the quadratic equation, (2x−1)(x−1)=0. Applying Definition 1.5.3 would give you:
  1. 2x2−3x+1=0
  2. (2x+1)=0 and (x+1)=0
  3. (2x−1)=0 and (x−1)=0
Answer.
C
🔗
(c)
🔗
🔗
After applying the zero product property, what are the solutions to the quadratic equation (2x−1)(x−1)=0?
  1. x=−12 and x=1
  2. x=12 and x=−1
  3. x=−12 and x=−1
  4. x=12 and x=1
Answer.
D
🔗

Remark 1.5.5.

🔗
Notice in Activity 1.5.2 and Activity 1.5.4, that not all equations are set up "nicely." You will need to do some manipulation to get everything on one side (AND in factored form!) and 0 on the other *before* applying the zero product property.
🔗

Activity 1.5.6.

🔗
Suppose you want to solve the equation 2x2+5x−12=0, which is NOT in factored form.
🔗
(a)
🔗
🔗
Which of the following is the correct factored form of 2x2+5x−12=0?
  1. (2x−3)(x−4)=0
  2. (2x+3)(x−4)=0
  3. (2x+3)(x+4)=0
  4. (2x−3)(x+4)=0
Answer.
D
🔗
(b)
🔗
🔗
After applying Definition 1.5.3, which of the following will be a solution to 2x2+5x−12=0?
  1. x=−32 and x=−4
  2. x=32 and x=4
  3. x=−32 and x=4
  4. x=32 and x=−4
Answer.
D
🔗

Activity 1.5.7.

🔗
Solve each of the following quadratic equations:
🔗
(a)
(2x−5)(x+7)=0
Answer.
x=52 and x=−7
🔗
(b)
3x(4x−1)=0
Answer.
x=0 and x=14
🔗
(c)
3x2−14x−5=0
Answer.
x=−13 and x=5
🔗
(d)
6−x2=5x
Answer.
x=1 and x=−6
🔗

Activity 1.5.8.

🔗
Suppose you are given the equation, x2=9:
🔗
(a)
🔗
🔗
How many solutions does this equation have?
  1. 0
  2. 1
  3. 2
  4. 3
Answer.
C
🔗
(b)
🔗
🔗
What are the solutions to this equation?
  1. x=0
  2. x=3
  3. x=9,−9
  4. x=3,−3
Answer.
D
🔗
(c)
🔗
How is this quadratic equation different than the equations we’ve solved thus far?
Answer.
When solving for x, we have to consider both the positive and negative values.
🔗

Definition 1.5.9.

🔗
🔗
The square root property states that a quadratic equation of the form
x2=k2
🔗
(where k is a nonzero number) will give solutions x=k and x=−k.
🔗
In other words, if we have an equation with a perfect square on one side and a number on the other side, we can take the square root of both sides to solve the equation.
🔗

Activity 1.5.10.

🔗
Suppose you are given the equation, 3x2−8=4:
🔗
(a)
🔗
🔗
What would be the first step in solving 3x2−8=4?
  1. Divide by 3 on both sides
  2. Subtract 4 on both sides
  3. Add 8 on both sides
  4. Multiply by 3 on both sides
Answer.
C
🔗
(b)
🔗
Isolate the x2 term and apply Definition 1.5.9 to solve for x.
Answer.
To continue to isolate the x2 term, the next step is to divide by 3 and then take the square root of both sides.
🔗
(c)
🔗
🔗
What are the solution(s) to 3x2−8=4?
  1. x=6,−6
  2. x=2,−2
  3. x=0
  4. x=2
Answer.
B
🔗

Activity 1.5.11.

🔗
Solve the following quadratic equations by applying the square root property (Definition 1.5.9).
🔗
(a)
3x2+1=28
Answer.
x=3,−3
🔗
(b)
5x2+7=47
Answer.
x=8,−8 or x≈−2.828,2.828
🔗
(c)
2x2=−144
Hint.
Recall that when you have a negative number under a square root, that gives an imaginary number (−1=i).
Answer.
x=72i,−72i or in simplified form: x=62i,−62i
🔗
(d)
(x+2)2+3=19
Hint.
Isolate the binomial (x+2) first.
Answer.
x=2,−6
🔗
(e)
3(x−4)2=15
Answer.
x=4+5,4−5
🔗

Remark 1.5.12.

Not all quadratic equations can be factored or can be solved by using the square root property. In the next few activities, we will learn two additional methods in solving quadratics.
🔗

Definition 1.5.13.

🔗
Another method for solving a quadratic equation is known as completing the square. With this method, we add or subtract terms to both sides of an equation until we have a perfect square trinomial on one side of the equal sign and a constant on the other side. We then apply the square root property.
🔗
Note: A perfect square trinomial is a trinomial that can be factored into a binomial squared. For example, x2+4x+4 is a perfect square trinomial because it can be factored into (x+2)(x+2) or (x+2)2.
🔗

Activity 1.5.14.

🔗
Let’s work through an example together to solve x2+6x=4. (Notice that the methods of factoring and the square root property do not work with this equation.)
🔗
(a)
🔗
🔗
In order to apply Definition 1.5.13, we first need to have a perfect square trinomial on one side of the equal sign. Which of the following number(s) could we add to the left side of the equation to create a perfect square trinomial?
  1. 4
  2. 9
  3. −9
  4. 2
Answer.
B
🔗
(b)
🔗
🔗
Add your answer from part a to the right side of the equation as well (i.e. whatever you do to one side of an equation you must do to the other side too!) and then factor the perfect square trinomial on the left side. Which equation best represents the equation now?
  1. (x+3)2=−5
  2. (x−3)2=13
  3. (x+3)2=13
  4. (x−3)2=−5
Answer.
C
🔗
(c)
🔗
🔗
Apply the square root property (Definition 1.5.9) to both sides of the equation to determine the solution(s). Which of the following is the solution(s) of x2+6x=4?
  1. 3+13 and 3−13
  2. −3+13 and −3−13
  3. 3+5 and 3−5
  4. −3+5 and −3−5
Answer.
B
🔗

Remark 1.5.15.

🔗
To complete the square, the leading coefficient, a (i.e., the coefficient of the x2 term), must equal 1. If it does not, then factor the entire equation by a and then follow similar steps as in Activity 1.5.14.
🔗

Activity 1.5.16.

🔗
Let’s solve the equation 2x2+8x−6=0 by completing the square.
🔗
(a)
🔗
Rewrite the equation so that all the terms with the variable x is on one side of the equation and a constant is on the other.
Answer.
2x2+8x=6
🔗
(b)
🔗
Notice that the coefficient of the x2 term is not 1. What could we factor the left side of the equation by so that the coefficient of the x2 is 1?
Answer.
Factor the left side by 2.
🔗
(c)
🔗
🔗
Once you factor the left side, what equation represents the equation you now have?
  1. 2(x2−8x)=−6
  2. 2(x2−4x)=−6
  3. 2(x2+4x)=6
  4. 2(x2+8x)=6
Answer.
C
🔗
(d)
🔗
🔗
Just like in Activity 1.5.14, let’s now try and create the perfect square trinomial (inside the parentheses) on the left side of the equation. Which of the following number(s) could we add to the left side of the equation to create a perfect square trinomial?
  1. 4
  2. 8
  3. −8
  4. 2
Answer.
A
🔗
(e)
🔗
🔗
What would we need to add to the right-hand side of the equation to keep the equation balanced?
  1. 4
  2. 8
  3. −8
  4. 2
Answer.
B. Students may need help understanding that adding 4 to the right-hand side of the equation does not keep the equation balanced.
🔗
(f)
🔗
🔗
Which of the following equation represents the quadratic equation you have now?
  1. 2(x+2)2=9
  2. 2(x−2)2=2
  3. 2(x+2)2=14
  4. 2(x−2)2=14
Answer.
C
🔗
(g)
🔗
Apply the square root property and solve the quadratic equation.
Answer.
x=−2+7,−2−7
🔗

Activity 1.5.17.

🔗
Solve the following quadratic equations by completing the square.
🔗
(a)
x2−12x=−11
Answer.
x=1,11
🔗
(b)
x2+2x−33=0
Answer.
x=−1+34,−1−34
🔗
(c)
5x2+29x=6
Answer.
x=−6,15
🔗

Definition 1.5.18.

The last method for solving quadratic equations is the quadratic formula - a formula that will solve all quadratic equations!
🔗
🔗
A quadratic equation of the form ax2+bx+c=0 can be solved by the quadratic formula:
x=−b±b2−4ac2a
🔗
where a, b, and c are real numbers and a≠0.
🔗

Activity 1.5.19.

🔗
Use the quadratic formula (Definition 1.5.18) to solve x2+4x=−3.
🔗
(a)
🔗
When written in standard form, what are the values of a, b, and c?
Hint.
Rewrite the equation so that your a-value is positive.
Answer.
a=1,b=4, and c=3. Students may also get a=−1,b=−4, and c=−3 if they moved x2 and 4x to the right side.
🔗
(b)
🔗
🔗
When applying the quadratic formula, what would you get when you substitute a, b, and c?
  1. x=−4±42−4(1)(3)2(1)
  2. x=4±(−4)2−4(1)(3)2(1)
  3. x=−4±(−4)2−4(1)(−3)2(1)
  4. x=4±42−4(1)(−3)2(1)
Answer.
A
🔗
(c)
🔗
🔗
What is the solution(s) to x2+4x=−3?
  1. x=−1,3
  2. x=1,3
  3. x=−1,−3
  4. x=1,−3
Answer.
C
🔗

Activity 1.5.20.

🔗
Use the quadratic formula (Definition 1.5.18) to solve 2x2−13=7x.
🔗
(a)
🔗
When written in standard form, what are the values of a, b, and c?
Hint.
Rewrite the equation so that your a-value is positive.
Answer.
a=2,b=−7, and c=−13. Students may also get a=−2,b=7, and c=13 if they moved 2x2 and −13 to the right side.
🔗
(b)
🔗
🔗
When applying the quadratic formula, what would you get when you substitute a, b, and c?
  1. x=−7±72−4(2)(13)2(1)
  2. x=7±(−7)2−4(2)(−13)2(2)
  3. x=−7±(−7)2−4(1)(−13)2(1)
  4. x=7±72−4(2)(−13)2(2)
Answer.
B
🔗
(c)
🔗
🔗
What is the solution(s) to 2x2−13=7x?
  1. x=7+734 and 7−734
  2. x=7+1534 and 7−1534
  3. x=−7+554 and −7−554
  4. x=−7+1554 and −7−1554
Answer.
B
🔗

Activity 1.5.21.

🔗
Use the quadratic formula (Definition 1.5.18) to solve x2=6x−12.
🔗
(a)
🔗
When written in standard form, what are the values of a, b, and c?
Hint.
Rewrite the equation so that your a-value is positive.
Answer.
a=1,b=−6, and c=12. Students may also get a=−1,b=6, and c=−12 if they moved x2 to the right side.
🔗
(b)
🔗
🔗
When applying the quadratic formula, what would you get when you substitute a, b, and c?
  1. x=−6±62−4(1)(12)2(1)
  2. x=6±(−6)2−4(1)(12)2(1)
  3. x=−6±(−6)2−4(1)(−12)2(1)
  4. x=6±62−4(1)(−12)2(1)
Answer.
B
🔗
(c)
🔗
🔗
Notice that the number under the square root is a negative. Recall that when you have a negative number under a square root, that gives an imaginary number (−1=i). What is the solution(s) to x2=6x−12?
  1. x=3+i3 and 3−i3
  2. x=6+i12 and 6−i12
  3. x=−3+i3 and −3−i3
  4. x=−6+i12 and −6−i12
Answer.
A
🔗

Activity 1.5.22.

🔗
Solve the following quadratic equations by applying the quadratic formula (Definition 1.5.18).
🔗
(a)
2x2−3x=5
Answer.
x=−1,52
🔗
(b)
4x2−1=−8x
Answer.
x=−2+52,−2−52
🔗
(c)
2x2−7x−13=−10
Answer.
x=7+734,7−734
🔗
(d)
x2−6x+12=0
Answer.
x=3+i3,3−i3
🔗

Activity 1.5.23.

🔗
Now that you have seen all the different ways to solve a quadratic equation, you will need to know WHEN to use which method. Are some methods better than others?
🔗
(a)
🔗
🔗
Which is the best method to use to solve 5x2=80?
  1. Factoring and Zero Product Property
  2. Square Root Property
  3. Completing the Square
  4. Quadratic Formula
Answer.
B
🔗
(b)
🔗
🔗
Which is the best method to use to solve 5x2+9x=−4?
  1. Factoring and Zero Product Property
  2. Square Root Property
  3. Completing the Square
  4. Quadratic Formula
Answer.
C or D
🔗
(c)
🔗
🔗
Which is the best method to use to solve 3x2+9x=0?
  1. Factoring and Zero Product Property
  2. Square Root Property
  3. Completing the Square
  4. Quadratic Formula
Answer.
A
🔗
(d)
🔗
Go back to parts a, b, and c and solve each of the quadratic equations. Would you still use the same method?
Answer.
Answers will vary. It is important, though, for students to look at the overall structure to help them determine which method might be the best method to use and then discuss their reasons why with their peers.
🔗

Exercises 1.5.2 Exercises

PrevTopNext