Solving Quadratic Equations
While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solving
quadratic equations requires much more than simply isolating the variable, as is required in solving linear
equations. This piece will outline the different types of quadratic equations, strategies for solving each
type, as well as other methods of solutions such as Completing the Square and using the Quadratic
Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression are needed
for this piece. Let’s take a look!
Standard Form of a Quadratic Equation
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎
Where 𝒂, 𝒃, 𝒂𝒏𝒅 𝒄 are integers
and 𝒂 ≥ 1
To solve an equation in the form 𝒂𝒙𝟐 + 𝒄 = 𝒌, for some value 𝒌. This is the simplest quadratic
I.
equation to solve, because the middle term is missing.
Strategy: To isolate the square term and then take the square root of both sides.
Ex. 1)
2𝑥 2 = 40
2𝑥 2
40
=
2
2
Isolate the square term, divide both sides by 2
𝑥 2 = 20
√𝑥 2 = √20
Take the square root of both sides
𝑥 = ± √20
Remember there are two possible solutions
𝑥 = ± 2√5
Simplify radical; Solutions
(Please refer to previous instructional materials Simplifying Radical Expressions )
II.
To solve a quadratic equation arranged in the form 𝒂𝒙𝟐 + 𝒃𝒙 = 𝟎.
Strategy: To factor the binomial using the greatest common factor (GCF), set the monomial
factor and the binomial factor equal to zero, and solve.
Ex. 2)
12𝑥 2 − 18𝑥 = 0
6𝑥(2𝑥 − 3) = 0
Factor using the GCF
6𝑥 = 0
Set the monomial and binomial equal to zero
𝑥=0

2𝑥 − 3 = 0
𝑥=
3
2
Solutions
In some cases, the GCF is simply the variable with coefficient of 1.
1
Solving Quadratic Equations
III.
To solve an equation in the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, where the trinomial is a perfect square.
This too is a simple quadratic equation to solve, because it factors into the form 𝑚2 = 0, for
some binomial 𝑚.
(For factoring instructional methods, select The Easy Way to Factor Trinomials )
Strategy: To factor the trinomial, set each binomial equal to zero, and solve.
Ex. 3) 𝑥 2 + 6𝑥 + 9 = 0
(𝑥 + 3)2 = 0
Factor as a perfect square
(𝑥 + 3)(𝑥 + 3) = 0
𝑥+3=0
𝑥+3=0
Set each binomial equal to zero
𝑥 = −3
𝑥 = −3
Solve
𝑥 = −3
IV.
Not necessary, but valuable step to show two solutions
Double root solution
To solve an equation in the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, where the trinomial is not a perfect
square, but factorable. Similar to the last example, this is a simple quadratic equation to solve,
because it factors into the form 𝑚𝑛 = 0, for some binomials 𝑚 and 𝑛.
Strategy: To factor the trinomial, set each binomial equal to zero, and solve.
Ex. 4)
2𝑥 2 − 𝑥 − 6 = 0

Using the factoring method from The Easy Way to Factor Trinomials, we need to
find two number that multiply to give 𝑎𝑐, or −12, and add to give 𝑏, or −1. These
values are −4 and 3. Rewrite the trinomial with these two values as coefficients to
𝑥 that add to the current middle term of −1𝑥.
2𝑥 2 − 4𝑥 + 3𝑥 − 6 = 0
Rewrite middle term
2𝑥 2 − 4𝑥 + 3𝑥 − 6 = 0
2𝑥(𝑥 − 2) + 3(𝑥 − 2) = 0
Factor by grouping
(𝑥 − 2)(2𝑥 + 3) = 0
Factor out the common binomial (𝑥 − 2)
𝑥−2=0
Set each binomial equal to zero
𝑥=2
V.
2𝑥 + 3 = 0
3
𝑥 = −2
Solutions
To solve a quadratic equation not arranged in the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, but factorable.
Strategy: To combine like terms to one side, set equal to zero, factor the trinomial, set each
binomial equal to zero, and solve.
2
Solving Quadratic Equations
Ex. 5)
6𝑥 2 + 2𝑥 − 3 = 9𝑥 + 2
−9𝑥
− 9𝑥
6𝑥 2 − 7𝑥 − 3 = 2
−2 − 2
6𝑥 2 − 7𝑥 − 5 = 0

To factor this trinomial, we are looking for two numbers that multiply to give 𝑎𝑐, or
−30, and add to give 𝑏, or −7. These values would be 3 and −10. Rewrite the
trinomial with these two values as coefficients to 𝑥 that add to the current middle
term of −7𝑥.
6𝑥 2 + 3𝑥 − 10𝑥 − 5 = 0
Rewrite middle term
6𝑥 2 + 3𝑥 − 10𝑥 − 5 = 0
3𝑥(2𝑥 + 1) − 5(2𝑥 + 1) = 0
Factor by grouping
*Careful factoring a −5 from the second group
(2𝑥 + 1)(3𝑥 − 5) = 0
Factor out the common binomial (2𝑥 + 1)
2𝑥 + 1 = 0
Set each binomial equal to zero
3𝑥 − 5 = 0
1
𝑥 = −2
𝑥=
5
3
Solutions
Now that we have explored some examples, I’d like to take this time to summarize the strategies used
thus far in solving quadratic equations. Keeping in mind the goal is to isolate the variable, the format of
the equation will dictate the strategy used to solve. When the quadratic does not have a middle term, a
term with a power of 1, it is best to first isolate the squared term, and then take the square root of both
sides. This essentially will result in two solutions of opposite values. For quadratics that do not have a
𝑐-value, arrange the equation so that 𝑎𝑥 2 + 𝑏𝑥 = 0, and then factor using the GCF. Set the monomial,
or the GCF, and the binomial equal to zero and solve.
When the quadratic has one or more 𝑎𝑥 2 ’s, 𝑏𝑥’s, and 𝑐’s, the like terms need to be combined to one
side of the equation and set equal to zero before determining if the trinomial can be factored. Once
factored, set each binomial equal to zero and solve. Keep in mind while combining like terms that 𝑎 must
be an integer greater than or equal to 1. The solutions to cases such as these may result in a double root
solution, found when the trinomial is factored as a perfect square, or two unique solutions, found when
the trinomial is factored into two unique binomials.
3
Solving Quadratic Equations
There may be other cases where a GCF can be factored out of the trinomial before factoring occurs. Since
this unit is focused on solving quadratic equations, the GCF would simply be a constant. The next
example to illustrates while it’s helpful to factor out the GCF before factoring the trinomial, it is not
imperative to do so and has no impact on the solution of the quadratic equation.
VI.
To solve a quadratic equation in which there is a GCF among the terms of a trinomial.
Strategy (A : To determine the GCF between the terms of the trinomial once it is in standard
form, factor out the GCF, factor the trinomial, set each binomial equal to zero, and
then solve.
Ex. 6A) 12𝑥 2 − 22𝑥 + 6 = 0
2(6𝑥 2 − 11𝑥 + 3) = 0

To factor this trinomial, we are looking for two numbers that multiply to give 𝑎𝑐, or
18, and add to give 𝑏, or −11. These values would be −9 and −2. Rewrite the
trinomial with these two values as coefficients to 𝑥 that add to the current middle
term of −11𝑥.
2(6𝑥 2 − 9𝑥 − 2𝑥 + 3) = 0
Factor out the GCF of 2 from each term
2[3𝑥(2𝑥 − 3) − 1(2𝑥 − 3)] = 0
Factor by grouping
2(2𝑥 − 3)(3𝑥 − 1) = 0
Factor out the common binomial (2𝑥 − 3)
2𝑥 − 3 = 0
Set each binomial equal to zero
𝑥=
3𝑥 − 1 = 0
3
2
𝑥=
1
3
Solutions
Strategy (B): To factor the trinomial, set each binomial equal to zero, and solve.
Ex. 6B) 12𝑥 2 − 22𝑥 + 6 = 0

To factor this trinomial, we are looking for two numbers that multiply to give 𝑎𝑐, or
72, and add to give 𝑏, or −22. These values would be −18 and −4. Rewrite the
trinomial with these two values as coefficients to 𝑥 that add to the current middle
term of −22𝑥.
12𝑥 2 − 18𝑥 − 4𝑥 + 6 = 0
6𝑥(2𝑥 − 3) − 2(2𝑥 − 3) = 0
Factor by grouping
(2𝑥 − 3)(6𝑥 − 2) = 0
Factor out the common binomial (2𝑥 − 3)
2𝑥 − 3 = 0
Set each binomial equal to zero
6𝑥 − 2 = 0
4
Solving Quadratic Equations
𝑥=

3
2
𝑥=
2
6
=
1
3
Solutions
Notice in Ex 6A, since the GCF did not have a variable. The purpose of factoring and
setting each binomial equal to zero is to solve for the possible value(s) for the
variable that result in a zero product. If the GCF does not have a variable, it is not
possible for it to make a product of zero. With that said, in later topics there will be
cases where a GCF will include a variable, leaving a factorable trinomial. This type of
case results in a possibility of three solutions for the variable, as seen in the example
below.
3𝑥(𝑥 2 + 5𝑥 + 6) = 0
3𝑥(𝑥 + 2)(𝑥 + 3) = 0
3𝑥 = 0
𝑥=0
𝑥+2=0
𝑥 = −2
𝑥+3=0
𝑥 = −3
At this point we need to transition to solving quadratics equations that do not have trinomials that are
factorable. To solve these types of equations, we have two options, (1) to Complete the Square, and (2)
to use the Quadratic Formula. Essentially, these two methods yield the same solution when left in
simplified radical form. For the remainder of this unit I will do the following:
VII.

Explain how to Complete the Square

Provide examples utilizing the Completing the Square method

Prove the Quadratic Formula starting with Completing the Square

Provide examples solving equations using the Quadratic Formula

Provide an example that parallels all three methods in this unit

Provide instructional strategies for solving quadratic equations
How to Complete the Square
Goal: To get (𝑥 ± 𝑚)2 = 𝑘 , where 𝑚 and 𝑘 are real numbers and 𝑘 ≥ 0
For equations that are not factorable and in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎 = 1,
1. Move constant term to the side opposite the variable 𝑥.
2. Take
1
2
of 𝑏 and square the result.
3. Add this term to both sides.
4. Create your perfect square set equal to some constant value 𝑘 ≥ 0.
5
Solving Quadratic Equations
VIII.
To solve quadratic equations using the Completing the Square method.
Ex. 7)
𝑥 2 + 6𝑥 − 5 = 0

Since there are no two integers that multiply to give 𝑎𝑐, or −5, and add to
give 𝑏, or 6, this trinomial is not factorable, and therefore, Completing the
Square must be used to solve for 𝑥.
𝑥 2 + 6𝑥 + _____ = 5 + _____
6 2
6 2
Move constant to the right
1
𝑏,
2
𝑥 2 + 6𝑥 + ( 2 ) = 5 + ( 2 )
Take
𝑥 2 + 6𝑥 + 9 = 14
Simplify
(𝑥 + 3)2 = 14
Factor trinomial as a perfect square
√(𝑥 + 3)2 = √14
Take the square root of both sides
𝑥 + 3 = ± √14
Simplify
𝑥 = −3 ± √14
Solve for 𝑥; Solutions
Ex. 8)

square it and add it to both sides
2𝑥 2 + 16𝑥 = 4
Before proceeding with Completing the Square, notice 𝑎 ≠ 1 and the constant term is
already on the opposite side of the variable terms. First step must be to divide both sides
of the equation by 2.
𝑥 2 + 8𝑥 = 2
Result after division by 2
𝑥 2 + 8𝑥 + _____ = 2 + _____
Preparation for Completing the Square
8 2
8 2
1
𝑥 2 + 8𝑥 + ( 2 ) = 2 + ( 2 )
Take 2 𝑏, square it and add it to both sides
𝑥 2 + 8𝑥 + 16 = 18
Simplify
(𝑥 + 4)2 = 18
Factor trinomial as a perfect square
√(𝑥 + 4)2 = √18
Take the square root of both sides
𝑥 + 4 = ± 3√2
𝑥 = −4 ± 3√2
Simplify
Solve for 𝑥; Solutions
At any point during the solving process, if a negative value exists under the radical, there will be NO REAL
SOLUTION to the equation. These types of equations will be explored later once the imaginary number
system has been learned.
6
Solving Quadratic Equations
IX.
Quadratic Formula
The Quadratic Formula is another method to solving a quadratic equation. Let’s take a look at how the
standard form of a quadratic equation can be transformed into the Quadratic Formula using the
Completing the Square method.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Standard Form of a quadratic equation
𝑎𝑥 2 𝑏𝑥
𝑐
0
+
+
=
𝑎
𝑎
𝑎
𝑎
𝑏
𝑐
𝑥2 +
𝑥+
= 0
𝑎
𝑎
𝑏
𝑐
𝑥2 +
𝑥= −
𝑎
𝑎
Ensure a coefficient of 1 for 𝑥 2 by dividing by 𝑎,
and move the constant term to the right

The square of half of what is now the 𝑏 term, or the middle term, is
1 𝑏 2
𝑏 2
𝒃𝟐
( ∙ ) = ( ) =
2 𝑎
2𝑎
𝟒𝒂𝟐
𝑥2 +
𝑏
𝒃𝟐
𝑐
𝒃𝟐
𝑥+
=
−
+
𝑎
𝟒𝒂𝟐
𝑎
𝟒𝒂𝟐
𝑥2 +
𝑏
𝑏2
4𝑎𝑐
𝑏2
𝑥+
=
−
+
𝑎
4𝑎2
4𝑎2
4𝑎2
Complete the Square
Get common denominator on the right
𝑏
𝑏2
−4𝑎𝑐 + 𝑏 2
𝑥 +
𝑥+
=
𝑎
4𝑎2
4𝑎2
2
(𝑥 +
𝑏 2
−4𝑎𝑐 + 𝑏 2
) =
2𝑎
4𝑎2
√(𝑥 +
Factor trinomial as a perfect square
𝑏 2
−4𝑎𝑐 + 𝑏 2
) = √
2𝑎
4𝑎2
𝑥+
𝑏
√−4𝑎𝑐 + 𝑏 2
= ±
2𝑎
2𝑎
𝑥=
−𝑏
√−4𝑎𝑐 + 𝑏 2
±
2𝑎
2𝑎
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
Take the square root of both sides
Simplify
Solve for 𝑥
Quadratic Formula
7
Solving Quadratic Equations
X.
To solve quadratic equations using the Quadratic Formula.
Ex 9.)
2𝑥 2 − 8𝑥 + 5 = 0
𝑎=2
𝑏 = −8
𝑐=5
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
𝑥=
−(−8) ± √(−8)2 − 4(2)(5)
2(2)
Substitute
𝑥=
8 ± √64 – 40
4
Evaluate
𝑥=
8 ± √24
4
𝑥=
8 ± 2√6
4
𝑥=
4 ± √6
2
Subtract
Simplify radical
Simplify fraction; Solutions
Ex. 10) 2𝑥 = 5 − 4𝑥 2
 Notice this equation is not in the standard form for quadratic equations.
Before identifying the values for 𝑎, 𝑏 and 𝑐, the equation must be arranged in
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 form. After adding 4𝑥 2 and subtracting 5, we get
4𝑥 2 + 2𝑥 − 5 = 0
𝑎 = 4 𝑏 = 2 𝑐 = −5
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
𝑥=
−(2) ± √(2)2 − 4(4)(−5)
2(4)
Substitute
𝑥=
−2 ± √4 + 80
8
Evaluate
𝑥=
−2 ± √84
8
𝑥=
−2 ± 2√21
8
Add
Simplify
8
Solving Quadratic Equations
𝑥=
−1 ± √21
4
Simplify fraction; Solution
As in Completing the Square, if a negative value results under the radical, there’s NO REAL SOLUTION.
XI.
Compare all three methods learned
Factoring
Completing the Square
Ex. 11) 4𝑥 2 − 8𝑥 − 5 = 0
Ex. 11) 4𝑥 2 − 8𝑥 − 5 = 0
 Two integers that multiply to
give −20 that add to give
−8 are −10 and 2.
 First step is to obtain a
coefficient of 1 for the 𝑥 2 by
dividing both sides of the
equation by 4.
4𝑥 2 − 10𝑥 + 2𝑥 − 5 = 0
2𝑥(2𝑥 − 5) + 1(2𝑥 − 5) = 0
(2𝑥 − 5)(2𝑥 + 1) = 0
2𝑥 − 5 = 0
5
𝑥=
2
2𝑥 + 1 = 0
1
𝑥= −
2
5
0
=
4
4
5
𝑥 2 − 2𝑥 − = 0
4
5
𝑥 2 − 2𝑥 =
4
𝑥 2 − 2𝑥 −
5
4
𝑥 2 − 2𝑥 + _____= + _____
2 2 5
2 2
𝑥 2 − 2𝑥 + ( ) = + ( )
2
4
2
5
𝑥 2 − 2𝑥 + 1 = + 1
4
9
𝑥 2 − 2𝑥 + 1 =
4
9
(𝑥 − 1)2 =
4
9
√(𝑥 − 1)2 = √
4
3
2
3
𝑥=1 ±
2
5
1
𝑥=
𝑥= −
2
2
𝑥−1= ±
9
Quadratic Formula
Ex. 11.) 4𝑥 2 − 8𝑥 − 5 = 0
𝑎=4
𝑥=
𝑥=
𝑏 = −8
𝑐 = −5
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
−(−8) ± √(−8)2 − 4(4)(−5)
2(4)
𝑥=
8 ± √64 + 80
8
8 ± √144
8
8 ± 12
𝑥=
8
20
4
𝑥=
𝑥= −
8
8
5
1
𝑥=
𝑥= −
2
2
𝑥=
Solving Quadratic Equations
XII.
Instructional Strategies
This is such a wonderful unit that builds on the familiar skills like solving equations, while setting up the
transition to exploring the graphical nature of quadratic solutions. Check out Being Strategic in Solving
Equations Part I & II to learn more about the flexibility in equation solving. Students have quite a bit of
flexibility in solving quadratic equations as well. This unit follows the factoring lessons in most
curriculums very closely. Essentially, the only new material in this unit is the Completing the Square and
the Quadratic Formula. It is imperative that you teach this unit in a progressive nature as I have laid out
here, starting with what students are familiar with, adding one layer at a time to arrive at the more
complex equations as illustrated in Examples 7 – 10.
Throughout the beginning of this unit, pose questions to students such as
 Does the equation have a middle term, or does the equation have a 𝑏 term?
 Is the equation in standard quadratic form?
 Is there a greatest common factor?
 Is the trinomial factorable?
 Can the trinomial be factored as a perfect square?
 How many unique solutions does the equation have?
Encourage students to ask these questions back to you or other students as equations are solved in class.
This will cause students to slow down and think carefully about the type of equation they are solving.
With that said, there is usually more than one approach to solving most equations. Take for instance
Example 11. Even if the equation is factorable, the Completing the Square method and the Quadratic
formula can be used to solve the equation; however, it may not be the most efficient method. Often
students will gravitate towards the formula because they are comfortable with mindless substitution and
computation that’s involved with a formula. Needless to say, they quickly realize they must be meticulous
weaving in and out of the steps so not to lose a sign or simplify incorrectly. In many cases, taking the
scenic route, or the more elaborate method of solution, will cause careless errors throughout the solving
process. The goal is for student to learn the process of examining what they have been given and proceed
with the method of solution that makes sense for the given equation.
To encourage this type of analysis and discourse, provide opportunities for students to showcase these
skills. One activity is to group students in 3’s, provide them with a quadratic equation to solve, have each
student demonstrate one of the methods of solution, and then decide as a group which method was the
most efficient or strategic. When presenting to the class, have each student explain why their method
was, or was not the most efficient. In a class, this could be 10 or more equations solved. Don’t shy away
from including equations that are missing terms or equations that are not in standard form. These might
prove to be more difficult, since they are required to think more carefully about what they have been
given, but they are very valuable learning tools. Following this activity, provide students with an
equation, and without requiring them to solve using paper and pencil, have them explain, either verbally
or in written form, which method they think would be the most strategic or most efficient. Keep in mind,
there is room for opinion in these responses. Simply listen and evaluate students thought process as they
explain. Skills such as these are invaluable and will help create well rounded mathematical thinkers.
10