Worksheet 6
SOLVING QUADRATIC EQUATIONS
Name: ___________________________________________
Warm Up Exercises
Evaluate the following:
a) 52 = _______
b) (0.2)2 = _______
c) (𝑥 + 1)2 = _____________________
Taking the Square Root
Recall that “squaring” a number refers to taking that number to the second power. For example,
we would say “three squared” to describe 32 = 9. Just as the “inverse” of multiplication is
division, the inverse of “squaring” a number is “square-rooting” a number, denoted by
which is called the “radical” symbol.
Example 1: Evaluate √25 and √0.04.
Solution:
√25 = √52 = 5
√0.04 = √(0.2)2 = 0.2
Remark: Note that the square root of any negative number does not exist because no number
times itself is negative. Therefore, you cannot take the square root of negative numbers.
Exercise 1
1) The 40 is in between what two consecutive positive integers?
2) The 160 is in between what two consecutive positive integers?
Forms of a Quadratic Equation: Standard versus Vertex Form
Quadratic equations are usually written in one of two forms: standard form or vertex form. While
vertex form (as we shall see in the next Activity) is primarily used for graphing, standard form is
most often used when solving quadratic equations and will be the focus of this lesson. Why is
putting a quadratic equation in standard form important? There are four primary methods of
solving quadratic equations, and three of the methods start with the quadratic equation in
standard form. Hence, it simply gives a starting point to solving a quadratic equation.
Standard Form:
Standard form of a quadratic equation is written as ax 2  bx  c  0 , where a, b, and c are
real numbers and a  0 .
Example 2: Write the following in standard form: x 2  2 x  7 .
Solution: The objective here is to set the equation equal to zero. Here, let’s subtract 2x
and 7 from both sides.
x 2  2x  7 .
x 2  2x  7  2x  7  2x  7
x 2  2x  7  0
Example 3: Write the following in standard form:  5 z 2  11  3z .
Solution: All we need to do here is add 3z from both sides of the equation.
 5 z 2  11  3z
 5 z 2  11  3z  3z  3z
 5 z 2  3 z  11  0
Vertex Form:
Vertex form of a quadratic equation is written as 𝑎(𝑥 − ℎ)2 + 𝑘 = 0, where 𝑎, ℎ,and 𝑘 are
real numbers and 𝑎 ≠ 0. This way of writing a quadratic equation is useful primarily for
graphing, but we have a convenient means of solving equations in this form.

Examples of equations in vertex form: (4 x  5) 2  49  0 , −2(𝑥 + 6) + 4 = 0
Exercise 2: Write the following in standard form.
a) (4 x  5) 2  49  0
b) ( x  3) 2  3( x  2)
Methods for Solving Quadratic Equations
Once a quadratic equation is put into standard form, we are ready to solve it. As mentioned
before, there are four primary methods for solving quadratic equations. Why do we need four?
We don’t. In fact, using the quadratic formula works EVERY time, but using this method is
sometimes slow and tedious, so the other methods are simply there to make our lives easier.
METHOD 1: FACTORING
Use this method if the quadratic expression in standard form is factorable.
The first (and most convenient) method to solve quadratic equations is by factoring which we
learned in the last activity. As you probably know by now, factoring quadratic expressions takes
some getting used to, but remember that not all quadratic expressions are factorable. Therefore,
this method only works in special cases, and if you cannot get the factoring to work out, then
move on to Method 2.
1. Put the equation in standard form.
2. Factor the quadratic expression.
3. Set each factor equal to zero and solve
Example 3: Solve: x 2  x  2  0
Solution: The equation is in already in the standard form so we move to Step 2. Factoring the
quadratic expression, we get
( x  2)( x  1)  0
Then we set each factor equal to zero and solve:
𝑥 + 2 = 0 OR x − 1 = 0
𝑥 = −2 OR
𝑥 = 1.
Remark: Why do we set each of the factors equal to zero? If you think of 𝑥 + 2 and 𝑥 − 1 as
unknown numbers, then we have two numbers multiplying to zero. The ONLY way two
numbers multiply to zero is if one of those numbers is zero. Therefore, 𝑥 + 2 = 0 or 𝑥 − 1 =
0.
Example 4: Solve: x( x  3)  2( x  2)
Solution: First we need to put the equation in the standard form. We do this by moving
everything to the left side of the equal sign and simplify.
𝑥(𝑥 − 3) − 2(𝑥 − 2) = 0
𝑥 2 − 3𝑥 − 2𝑥 + 4 = 0
𝑥 2 − 5𝑥 + 4 = 0
Now we factor and set each factor equal to zero and solve for x.
(𝑥 − 4)(𝑥 − 1) = 0
𝑥 − 4 = 0 𝑂𝑅 𝑥 − 1 = 0
𝑥 = 4 𝑂𝑅
𝑥=1
Exercise 3: Use the factoring method to solve the following.
a) x 2  6 x  8  0
b) x 2  11x  30
Warm-Up:
a)
156
b)
6300
METHOD 2: SQUARE ROOT METHOD
Use this method when the quadratic expression is similar to the form (𝑎𝑥 + 𝑏)2 = 𝑐.
This is the only method that does not require our quadratic equation to be in standard form.
As in the case of the factoring method, we can only use this method in special cases. The Square
Root Property states that for any real number c, if x 2  c , then x  c or x   c .
Example 5: Use the Square Root Property to solve: (2 x  5) 2  6  0
Solution: First, add 6 to both sides of the equal sign.
(2𝑥 + 5)2 = 6
2𝑥 + 5 = ±6
2𝑥 = −5 ± √6
Use the Square Root Property.
Solve for x by first subtracting 5 from both sides.
Divide both sides by 2.
𝑥=
So 𝑥 =
−5+√6
2
−5 ± √6
2
or 𝑥 =
−5−√6
2
Exercise 4: Use the Square Root Property to solve the following.
a) (3x  2) 2  15
b) (5 x  4) 2  9
2
3
9

c)  x   
2
4

d) ( 4  x) 2 
36
25
METHOD 3: COMPLETING THE SQUARE
Use this method for any quadratic equation. Unlike other methods that we have seen so far, this
method works EVERY time.
The idea underlying this method is to take a quadratic equation in standard form and
manipulate it into a form to which the Square Root Method can be applied. This method will also
be used in the next section to change a quadratic equation from standard form to vertex form.
The general steps to completing the square can be very confusing. Therefore, we will work
through a few examples and then provide a shortcut to save you time.
Example 6: Solve x 2  4 x  3  0 by completing the square.
Solution:
𝑥 2 + 4𝑥 + 3 = 0
𝑥 2 + 4𝑥 = −3
make left side a perfect square
Isolate the constant term
𝑏
2
Add (− 2𝑎) =4 to both sides to
𝑥 2 + 4𝑥 + 4 = −3 + 4
Note that 𝑥 2 + 4𝑥 + 4 = (𝑥 + 2)2
(𝑥 + 2)2 = 1
Now, we can solve this using the Square Root Method.
(𝑥 + 2)2 = 1
𝑥 + 2 = ±1
𝑥 = −2 ± 1
So 𝑥 = −3 or −1
Example 7: Solve 𝑥 2 − 6𝑥 − 8 = 0 by completing the square.
Solution:
𝑥 2 − 6𝑥 − 8 = 0
𝑥 2 − 6𝑥 = 8
Isolate the constant term
𝑏
2
Add (− 2𝑎) =9 to both sides to
make
left side a perfect square
𝑥 2 − 6𝑥 + 9 = 8 + 9
Note that 𝑥 2 + 4𝑥 + 4 = (𝑥 + 2)2
(𝑥 − 3)2 = 17
Now, we can solve this using the Square Root Method.
(𝑥 − 3)2 = 17
𝑥 − 3 = ±√17
𝑥 = 3 ± √17
Exercise 5:
Solve the following by completing the square. Show your work.
b) 2x2 - 5x – 7 =0
a) x2 -2x - 4= 0
Simplify, and show your work:
 10  6 2  4(2)(3)
c)
2(5)
 3  4 2  4(2)( 2)
d)
2(9)
METHOD 4: QUADRATIC FORMULA
Use this method for ANY quadratic equation.
If it looks like a quadratic equation can’t be solved by factoring or the square root method,
then Quadratic Formula can always be used and is often faster than completing the square. It’s a
bit tedious, but it always works. The quadratic formula states that,
If ax  bx  c  0 , then
2
 b  b 2  4ac
.
x
2a
Example 8: Solve 3x 2  4 x  6  0 by using the quadratic formula.
Solution: In order to use the quadratic formula, we need to know what our coefficients are.
In this case, a  3, b  4 and c  6 . Let’s substitute these values into the quadratic
formula.
x
 b  b 2  4ac
2a
 (4)  (4) 2  4(3)( 6)
x
2(3)
x
4  16  (72)
6
x
4  88
6
We can actually simplify this further. Note that
√88 = √4 ∗ 22 = √4 ∗ √22 = 2 ∗ √22 = 2√22.
Then
𝑥=
4 ± √88 4 ± 2 √22 2(2 ± √22) 2 ± √22
=
=
=
6
6
6
3
So,
𝑥=
2+√22
3
or 𝑥 =
2−√22
3
Exercise 6: Solve the following using the quadratic formula. Show your work.
a) x 2  6 x  9  0
b) x 2  7 x  10  0
c) 2 x 2  2  4 x
d) 2 x 2  7 x  9  0
OPTIONAL: Deriving the Quadratic Formula using Completing the Square
Given the general standard form of a quadratic equation, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, we can treat a, b,
and c just like normal numbers and solve this equation using completing the square. By doing
this, we obtain the quadratic formula which, as mentioned earlier, can be used to solve ANY
quadratic equation (since any quadratic equation can be written in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0).
This is a common theme in higher level mathematics. Instead of doing the same computation
over and over again, we often try to solve things in a general form such as this.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Isolate the constant term.
𝑎𝑥 2 + 𝑏𝑥 = −𝑐
𝑏
Divide both sides by a.
𝑐
𝑏
𝑥2 + 𝑎 𝑥 = − 𝑎
𝑏
2
Add (2𝑎) to both sides to complete the square.
𝑏
2
𝑐
𝑏
𝑏
2
𝑐
𝑏2
𝑏
2
𝑐
𝑏
2
4𝑎𝑐
𝑏
2
2
𝑥 2 + 𝑎 𝑥 + (2𝑎) = − 𝑎 + (2𝑎)
𝑏
𝑏
2
𝑏
Get a common denominator of 4𝑎2
(𝑥 + 2𝑎) = − 𝑎 + 4𝑎2
𝑏2
4𝑎
(𝑥 + 2𝑎) = − 𝑎 (4𝑎) + 4𝑎2
𝑏2
(𝑥 + 2𝑎) = − 4𝑎2 + 4𝑎2
(𝑥 + 2𝑎) =
𝑏 2 −4𝑎𝑐
𝑏 2 −4𝑎𝑐
𝑏
𝑥 + 2𝑎 = ±√
𝑏
𝑥 + 2𝑎 = ±
𝑏
𝑥 + 2𝑎 = ±
𝑏
𝑥 + 2𝑎 = ±
4𝑎2
√𝑏 2 −4𝑎𝑐
√4𝑎2
√𝑏 2 −4𝑎𝑐
√(2𝑎)2
√𝑏 2 −4𝑎𝑐
2𝑎
𝑏
𝑥 = − 2𝑎 ±
𝑥=
Now solve by Square Root Method
4𝑎2
√𝑏 2 −4𝑎𝑐
2𝑎
−𝑏±√𝑏 2 −4𝑎𝑐
2𝑎
2
Note 𝑥 2 + 𝑎 𝑥 + (2𝑎) = (𝑥 + 2𝑎)
Does this look familiar?