Up Your Score (39 page)

(2
x
+ 3) = 0 (
x
– 2) = 0

Now solve each equation for
x
. You should come up with two solutions:
x
can be either –
or 2.

This might have taken you a minute. It might have taken you twenty. But don’t worry about it—the more problems you do, the better you get at seeing the answer without having to do many calculations.

Not all trinomials can be factored like this, but more on that coming up.

Example: The sum of the two roots of a quadratic equation is 5 and their product is –6. Which of the following could be the equation?
(A)
x
²
– 6
x
+ 5 = 0
(B)
x
²
– 5
x
– 6 = 0
(C)
x
²
– 5
x
+ 6 = 0
(D)
x
²
+ 5
x
– 6 = 0
(E)
x
²
+ 5
x
+ 6 = 0

First of all, since the product is negative and the sum is positive, you know that one of the numbers is negative and the other is positive. Start with the product (there are fewer possibilities)—how many pairs of numbers would have a product of –6? The answer: –1 and 6, –6 and 1, –3 and 2, –2 and 3. Then look at the sum. Out of all of these pairs, which one has a sum of 5? –1 and 6 satisfy both conditions, so –1 and 6 are the roots.

Next, put those roots down as the answers to a quadratic equation—(
x
+ 1)(
x
– 6) = 0. Then FOIL this out, and you’ll get the answer: (B).

A quick tip: In any quadratic equation
ax
²
+
bx
+
c
= 0, the sum of the two roots is equal to –
and the product of the two roots is equal to
Try it, it works!

JaJa says: Watch your signs in these problems! The SAT will purposefully include answer choices that only differ by a + or − sign.

Sooner or later, you’ll try to solve a quadratic equation that looks something like this: 5
x
²
– 3
x
– 4 0. Sometime after the thirtieth try, you’ll realize that it doesn’t factor nicely. That’s where the quadratic formula comes in.

With a trinomial in the form a
x
²
+ b
x
+ c = 0

the two roots are
x =

(If you need to convince yourself, try this using a trinomial that you know factors nicely.)

Example: Solve for
x
if 5
x
²
– 3
x
– 4 = 0
In this trinomial, a = 5, b = –3, and c = –4.
Plug these into the formula.

So the two factors are

Since the quadratic formula works for every trinomial in the entire universe, it may be much easier for you to do if you don’t like the two-binomial thing you tried above.

When you get two linear equations, and you’re supposed to solve for
x
and
y
, there are two ways of tackling them. (Or three, if you have a graphing calculator.) Sometimes the first method is easier, and sometimes the second method is. Know how to use both of them.

Say that you’re supposed to solve for this:

2
x
+ 3
y
= 6
5
x
+ 8
y
= 11

Keep them stacked up, then multiply or divide the equations to make the coefficients of either the
x
’s or the
y
’s equal. Sometimes they’re equal when you start out. But this time we’re not so lucky. Let’s say we go with the
x
’s. In order to make the coefficients equal, multiply the top equation by 5 and the bottom equation by 2:

5(2
x
+ 3
y
) = (6)5
2(5
x
+ 8
y
) = (11)2

You end up with:

10
x
+ 15
y
= 30
10
x
+ 16
y
= 22

Subtract the bottom equation from the top one:

10
x
+ 15y = 30
–10
x
– 16
y
= –22
–
y
= 8
y
= –8

Then take any equation and plug in –8 for
y
.

2
x
+ 3
y
= 6
2
x
+ 3(–8) = 6
x
= 15

Or, you can take the first equation, and solve for
y
in terms of
x
(or solve for
x
in terms of
y
—it’s up to you).

Then you plug in the whole thing for the
y
in the second equation (if you solved for
x
in terms of
y
, then just plug in the whole thing for
x
).

Multiply both sides by 3 to get rid of the fraction.