NOTES: A quick overview of quadratic equations
LEVEL 1
What is a quadratic equation?
 Basically, it’s an equation with an x2 term in it, such as 2x2 – 3x + 1 = 0
 In its standard form it looks like this: ax2 ± bx ± c = 0
[See (*) note.]
 If we graphed y = 2x2 – 3x + 1 we’d get a parabola, which looks like a “U.”
A parabola can cross the horizontal axis
(the x-axis) either twice—as it does here—
or just once at the curviest part (the vertex)
or not at all.
When we set the equation = 0 we find where the
y-value is 0, that is, where the line crosses the
x-axis. Those places are called roots or zeros
or solutions. That’s what you’re finding when you
solve quadratic equations for x.
Here the solutions are x = 2 and x = 8.
What do the solutions of a quadratic equation mean?
 Quadratics are used to describe the path of objects, such as balls or missiles.
EXAMPLE: Suppose you threw a ball straight up from 2 feet off the ground at
46 feet per second (about 31 mph). After how many seconds would the ball land?
The height of the ball
h = –32t2 + 46t + 2
gravity
your toss
where t = time in seconds
height ball started
The ball will land when the height is 0, when it’s 0 feet off the ground.
When we set the equation equal to 0 and solve, we find out it takes about 1.5 sec.
Don’t worry! You won’t have to handle word problems like this right now. Just know
that quadratic equations model all sorts of everyday phenomena—not only the path of
projectiles but also business calculations for how to set a price to make the most
profit, and geometric applications, such as finding the maximum height under an
underpass or the width of a ball field with synthetic turf, both the underpass & the field
actually being parabolas.
* Technically, the standard form of a quadratic is ax2 + bx + c = 0 with the understanding that actual examples often
involve subtraction signs, subtraction being the same as adding a negative. I find it less confusing at this introductory
level just to say the form is ax2 ± bx ± c = 0.
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Which method should you use to solve a quadratic equation?
• There are various methods. The best one to use depends on a combination of
your particular skills and what the equation looks like.
• Regardless of the method, you can always check your answer(s) by plugging
back into the original equation—though sometimes it’s messy to do so.
• Note that whereas when you solved linear equations (for example, 2x = 12), the
solution was a single value, with quadratic equations you may get two, one, or no
real solutions. For quadratics we sometimes speak of a solution set in braces { }.
method
factoring
pros
cons
 can be quick, especially  works only with factorable
when coefficient of x is 1
equations
 uses guess-and-check, not
rote procedure
quadratic formula
 works with any quadratic  takes a long time
equation
 offers chances for careless
 definite, rote procedure
errors
finding roots
 quick & not too hard
 works only on cases
without an x term
(for example, x2 = 36
or x2 – 36 = 0)
completing the square  works with any quadratic  can get messy
equation
 can be hard to remember
How do you solve a quadratic equation by factoring?
 This method works only for factorable quadratic equations.
1. Make sure the given equation is in standard form:
LEVEL 2
ax2 ± bx ± c = 0
o If it isn’t in standard form yet, you’ll have to adjust it first.
2. Factor the trinomial if possible.
[See A quick overview of factoring pages 4-6.]
o Note that many (actually most) quadratics are NOT factorable.
3. Set each factor that has a variable in it equal to 0 and solve each one for x.
o There may be 2 solutions (if the parabola intersects the x-axis twice)
or one (if the parabola touches the x-axis just once at its vertex).
GED test tip: If a factoring/quadratic problem is multiple choice, it might be quicker
to use FOIL on each choice to see which gets you back to
trinomial.
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Quadratics 2
EXAMPLE 1: Solve the quadratic equation x2 – 5x – 6 = 0 with factoring
1. Is the equation in standard form? YES
2. Factor the trinomial (doing FOIL in reverse): (x – 6)(x + 1) = 0
3. What values of x would make that whole product (x – 6)(x + 1) come out 0?
Well, 0 times anything is 0 so if (x – 6) = 0, the whole product would = 0,
too. It wouldn’t matter what the value of (x + 1) was.
Likewise, if (x + 1) = 0, the whole product would = 0. It wouldn’t matter
what the value of (x – 6) was.
When would x – 6 = 0 ? when x = 6
And when would x + 1 = 0 ? when x = 1
There are 2 solutions: when x = 6 and when x = 1
solution set: {1, 6}
The graph intersects the horizontal axis at those two values of x.
You can check yourself by plugging each solution back into the equation and
seeing if you get a true statement.
Does 62 – 5(6) – 6 = 0 ?
YES
2
Does (1) – 5(1) – 6 = 0 ? Does 1 + 5 – 6 = 0 ?
YES
EXAMPLE 2:
[harder]
Solve the quadratic equation 2x2 + 11x = –5 with factoring
1. Is the equation in standard form? NO, not yet. It’s not equal to 0.
2x2 + 11x
= –5
+5
+5
2
2x + 11x + 5 = 0 Now it’s in standard form.
2. Factor the trinomial (doing FOIL in reverse): (2x + 1)(x + 5) = 0
3. Set each factor that has a variable in it equal to 0 and solve each one for x.
2x + 1 = 0
2x = –1
2x = –1
x=−
1
2
or
x+5=0
or
or
x = –5
x = –5
or
x = –5
1
2
solution set: {– 5 , − }
There are 2 solutions. The graph intersects the x-axis at those two values of x.
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How do you use the quadratic formula? LEVEL 1
 This method works for all quadratic equations.
 The quadratic formula is on your GED test formula sheet:
𝑥=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
1. Make sure the given equation is in standard form:
ax2 ± bx ± c = 0
o If it isn’t in standard form yet, you’ll have to adjust it first.
2. Write down the values for a, b, and c.
o If there’s a “–“ in front of b, then b is negative.
o If there’s a “–“ in front of c, then c is negative.
3. Plug the values for a, b, and c into the quadratic formula,
and crank out the order of operations until you’ve solved for x.
o –b means the opposite of b. If b = 5 then –b = –5. If b = –5 then –b = 5
o Be careful when finding b2 if you use your calculator: (–10)2  –102
 If b is negative, you need to use parentheses or remember that a
negative times a negative is positive.
o Usually you’ll get 2 possible values for x, but if the square root value
(the discriminant) = 0, there will be just 1 solution. At this level, we’re
dealing with real numbers, so the number inside √
won’t be negative.
EXAMPLE 1: Solve with the quadratic formula: x2 – 10x + 16 = 0
1. Is the equation in standard form? YES
2. What are the values of a, b, and c?
a=1
[If there’s no coefficient shown, it’s an implied 1.]
b = –10 [The “–“ goes with the variable. Subtraction means “plus a negative.”]
c = 16
3. Plug in and crank it out one step at a time.
𝑥=
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
10±√100−64
2
=
10±√(10)2 −4(1)(16)
2(1)
[You can’t cancel the 10 and 2 because there’s +/– on top.]
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=
=
10±√36
2
10±6
2
Now you split the solutions in two, one for the + and one for the  in the 
𝑥=
10+6 16
2 = 2 =8
or
𝑥=
10  6 4
2 = 2 =2
There are 2 possible solutions: x = 8 and x = 2.
The parabola crosses the horizontal axis twice.
solution set: {2, 8}
You can check your answer by plugging first one and then the other into the
original equation (x2 – 10x + 16 = 0) and seeing if it comes out true.
EXAMPLE 2: Solve with the quadratic formula: 18x2 + 36x – 13 = 19
1. Is the equation in standard form? NO, not yet. It’s not equal to 0.
18x2 + 36x – 13 = 19
–19 –19
2
18x + 36x – 32 = 0
Now it’s in standard form.
2. What are the values of a, b, and c?
a = 18
b = 36
c = –32
3. Plug in and crank it out one step at a time.
𝑥=
=
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
36 ± √3600
=
36 ± √(36)2 −4(18)(32)
2(18)
36
36 ± 60
36
Now you split the solutions in two, one for the + and one for the  in the 
𝑥=
36+ 60
36
=
24
36
=
There are 2 solutions: x =
2
3
2
3
or
𝑥=
and x = 
8
3
36 – 60
36
=
96
36
=
8
3
solution set: {
8 2
3,3}
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How do you solve a quadratic equation by finding roots?
 This method works only if there’s no x term, just an x2 term and a constant.
LEVEL 1
1. Isolate x2.
o Use the techniques you learned for solving linear equations to get x2
by itself on one side of the “=” and a lone number on the other.
2. Take the square root of each side.
o Note that √𝑥 2 = x.
o The side with the constant will be both the positive square root √
and the negative square root √ of the constant.
EXAMPLE 1: Solve for x: x2 – 16 = 0
1. Isolate the x2:
x2 – 16 = 0
+ 16 + 16
x2 = 16
2. Take the square root of each side.
x2 = 16
√𝑥 2 = ± √16
x = √16 = 4
or
x = –√16 = –4
solution set: {4, 4}
(the positive square root of 16 or the negative square root of 16)
EXAMPLE 2: Solve for x: 2x2 + 13 = 175
2x2 + 13
–13
=
175
–13
2x2 = 162
2𝑥 2 162
=
2
2
x2 = 81
x =  √81 = 9 or –9
solution set: {9, 9}
You can check yourself by plugging 9 in for x in the original equation
and making sure the result is 175. Then do the same with –9.
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EXAMPLE 2: Solve for x: x2 – 8 = 55
x2 – 8 = 55
+8
+8
x2 = 63
x =  √63
x =  3√7
x = 3√7 or –3√7
solution set: {–3√7 , 3√7}
How do you solve a quadratic by completing the square?
 This method works for all quadratic equations.
 Introductory remarks:
o From the method of finding squares, we know how to solve x2 = 25.
Taking the square root of both sides, we get x = 5 or x = –5
o With a little mental stretching, we see we could use the same technique to
solve (x + 3)2 = 25.
Taking the square root of both sides, we get x + 3 = 5 or x + 3 = –5
Solving these 1-step equations, we get x = 2 or x = –8
We can do the same for (x – 3)2 = 25.
Taking the square root of both sides, we get x – 3 = 5 or x – 3 = –5
Solving these 1-step equations, we get x = 8 or x = –2
x2 + 6x + 9 factors into (x + 3)2 and x2 – 6x + 9 factors into (x – 3)2
These are perfect square trinomials—They factor into perfect squares.
o If we could manipulate a quadratic equation so it had a perfect square
trinomial, we’d be in luck, and we could solve it pretty easily.
That’s basically what completing the square is all about—but it can get
messy with lots of fractions.

This is an advanced topic. If you’re interested, see me or watch the videos.
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