Quadratics
Quadratic equations are equations that can be written in standard form as y = ax 2 + bx + c,
where a, b, and c are constants. When graphed, a quadratic equation forms a parabola. For
the purposes of both the SAT and the ACT, you should be familiar with solving quadratics by
factoring and by graphing.
Typically, you should attempt to solve a quadratic by factoring if the a term is 1. If a quadratic
has an a term other than one, you should probably just solve it graphically and be done with it.
The one exception to this is when a quadratic has an a term that is a factor of the b and c terms,
in which case you can divide both sides of the equation by a, thereby reducing the a term to 1.
This method is demonstrated in example 4.
Solving by Factoring
To solve a quadratic by factoring, begin by putting the quadratic into standard form by setting
the entire equation equal to zero. Then find two numbers that add to the b term and multiply
to the c term. These numbers will allow you to write the factored form of the quadratic, which
will in turn allow you to figure out the solutions. Observe the following examples.
1) x2 + 7x + 12 = 0
To solve for x, find two numbers that add to 7 and multiply to 12. The two numbers that meet
these criteria are 3 and 4. Therefore, the factored form of this equation is (x + 3)(x + 4) = 0. To
make this equation true, the left side of the equation must equal 0, which happens when either
of the two factors is equal to zero. x + 3 = 0 when x = -3, and x + 4 = 0 when x = -4, so x = -3 and
x = -4 are the two solutions to this quadratic.
2) x2 – 2x – 24 = 0
To solve for x, find two numbers that add to -2 and multiply to -24. The two numbers that meet
these criteria are -6 and 4. Therefore, the factored form of this equation is (x – 6)(x + 4) = 0. To
make this equation true, the left side of the equation must equal 0, which happens when either
of the two factors is equal to zero. x – 6 = 0 when x = 6, and x + 4 = 0 when x = -4, so x = 6 and x
= -4 are the two solutions to this quadratic.
3) x2 = 9x – 18
Begin by getting the quadratic into standard form by setting it equal to zero. x 2 – 9x + 18 = 0.
Then, find two numbers that add to -9 and multiply to 18. The two numbers that meet these
criteria are -3 and -6. Therefore, the factored form of this equation is (x – 3)(x – 6) = 0. To
make this equation true, the left side of the equation must equal 0, which happens when either
of the two factors is equal to zero. x – 3 = 0 when x = 3, and x – 6 = 0 when x = 6, so x = 3 and x
= 6 are the two solutions to this quadratic.
4) 2x2 + 16x + 30 = 0
Begin by dividing both sides of the equation by 2 so that the a term is equal to 1. x2 + 8x + 15 =
0. Then, find two numbers that add to 8 and multiply to 15. The two numbers that meet these
criteria are 3 and 5. Therefore, the factored form of this equation is (x + 3)(x + 5) = 0. To make
this equation true, the left side of the equation must equal 0, which happens when either one
of the two factors is equal to zero. x + 3 = 0 when x = -3, and x + 5 = 0 when x = -5, so x = -3 and
x = -5 are the two solutions to this quadratic.
Expanding Quadratics: FOIL
You should also be familiar with how to move from the factored form of a quadratic to standard
form. To do so, you must expand the polynomial by multiplying the two factors together. This
is most easily done via the FOIL method. Multiply the two First terms, the two Outer terms, the
two Inner terms, and the two Last terms. Then simplify.
5) Write (x + 4)(x – 8) = 0 in standard form.
To expand, use the FOIL method described above. Multiply the two First terms, the two Outer
terms, the two Inner terms, and the two Last terms, and then simplify. x • x + x • -8 + 4 • x + 4
• -8 = x2 – 8x + 4x – 32. Combine like terms to get x2 – 4x – 32 = 0.
6) Expand (x + 7)2.
Avoid making the common mistake of thinking that (x + 7)2 = x2 + 72. Instead, (x + 7)2 = (x + 7)(x
+ 7). Using the foil method, this expression expands to x 2 + 7x + 7x + 49 = x2 + 14x + 49.
Because this expression was a perfect square (see below), you alternatively could have
expanded this expression without using FOIL by knowing the identity that (x + a)2 = x2 + 2ax + a2.
Solving Quadratics Graphically
Any quadratic that can be solved algebraically can also be solved graphically. Typically, if you
are proficient at solving quadratics via factoring, factoring is the quickest and easiest approach
on problems that are easy to factor. However, on problems that are difficult or impossible to
factor, graphing is the best approach. Graphing is also the best approach on all quadratics if
you are not proficient at factoring. To solve a quadratic graphically on your graphing calculator,
set the equation equal to zero, then enter it into the “y =” field on your graphing calculator.
Once you have it graphed, find the x- intercepts (the x-values where the graph crosses the xaxis). These x-intercepts represent the solutions to the quadratic. If the answer choices are
reasonably far from each other, you can use the trace function to get an approximate value of
the x-intercepts. If you need greater precision, use the CALC zero tool on a TI calculator,
since a zero is another term for an x-intercept.
7) 3x2 + 5x – 8 = 0
This is an ugly quadratic which could not be factored with integers. It is a perfect example of a
quadratic that is most easily solved graphically. Enter 3x 2 + 5y – 8 into the “y =” field on your
graphing calculator, and find the x- intercepts (the x-values where the graph crosses the x-axis).
Using the CALC zero tool on a TI calculator, you get that the two intercepts are -8/3 and 1.
Therefore, -8/3 and 1 are the two solutions.
Quadratic Formula
The quadratic formula allows you to plug in the a, b and c values from standard form, y = ax2 +
bx + c, to find the two solutions of a quadratic. The quadratic formula is as follows:
√
. If you are familiar with using the quadratic formula to find the solutions of a
quadratic, it is another good approach. If you are not familiar with this equation, you are
probably better off learning to solve quadratics graphically instead of memorizing this equation.
Quadratic Identities
The following quadratic identities are featured very frequently on both the SAT and the ACT. It
is worth memorizing these identities.
Perfect squares: (x + a)2 = x2 + 2ax + a2
(x – a)2 = x2 – 2ax + a2 (perfect squares are the
quadratics with exactly one solution)
Examples: (x + 5)2 = x2 + 10x + 25
(x – 4)2 = x2 + 8x + 16
Difference of two squares: (x + a) (x – a) = x2 – a2
Example: (x + 3) (x – 3) = x2 – 9
8) If x2 + 16x + k = 0 and there is only one possible solution for x, what is the value of k?
If there is only one possible solution for x, this quadratic must be a perfect square. Because the
b term is positive, it must be of the form (x + a)2 = x2 + 2ax + a2. Therefore 2a = 16, so a = 8. k =
a2, so k = 82 = 64.
9) If x + y = 8 and x2 – y2 = 24, what is the value of x – y?
When the SAT or ACT asks you to solve for an expression like “x – a” rather than just solving for
one variable or the other, it is very likely that you can solve for that expression without ever
finding the value of the individual variables. In this case, you can solve directly for x – a by using
the difference of two squares. The difference of two squares states (x + a) (x – a) = x2 – a2.
Therefore, x – a = (x2 – a2)/(x + a). In this problem, x – y = (x2 – y2)/(x + y) = 24/8 = 3.
Factoring to Simplify Rational Expressions
When you are asked to simplify a rational expression (an expression that contains a polynomial
in the numerator and a polynomial in the denominator), you should factor any quadratics and
look to cancel common factors between the numerator and the denominator.
10)
To solve this problem, factor both the numerator and denominator, and then cancel like
factors.
(
)(
)
(
)(
)
=
11)
To solve this problem, factor both the numerator and denominator, and then cancel like
factors.
(
)(
)
(
)(
)
=
12)
To solve this problem, factor both the numerator and denominator, and then cancel like
factors.
(
)(
(
)
)
=