FACTORABLE QUADRATIC EXPRESSIONS WITH ONE VARIABLE A quadratic expression includes a “squared” variable, such as x2.. An equation is quadratic if you can express it in this general form:
Trang 124. a + b = 30
2b = 60 2 2a
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (D) An unwary test taker might waste time trying to
find the values of a and b, because the centered data appears at first glance to
provide a system of two linear equations with two unknowns But you can easily
manipulate the second equation so that it is identical to the first:
2b = 60 2 2a
2b = 2~30 2 a!
b = 30 2 a
a + b = 30
As you can see, you’re really dealing with only one equation Since you cannot
solve one equation in two unknowns, you cannot make the comparison
Whenever you encounter a Quantitative Comparison question that calls for solving
one or more linear equations, stop in your tracks before taking pencil to paper Size up
the equation to see whether it’s one of the two unsolvable kinds you learned about
here If so, unless you’re given more information, the correct answer will be choice (D)
FACTORABLE QUADRATIC EXPRESSIONS WITH ONE
VARIABLE
A quadratic expression includes a “squared” variable, such as x2 An equation is
quadratic if you can express it in this general form:
ax2
1 bx 1 c 5 0
Where:
x is the variable
a, b, and c are constants (numbers)
a Þ 0
b can equal 0
c can equal 0
Here are four examples (notice that the b-term and c-term are not essential; in other
ALERT!
If the centered information in
a Quantitative Comparison consists of one or more linear
equations, never assume you
can solve for the variable(s).
Trang 2Quadratic Equation
Same Equation, but in the Form:
ax21 bx 1 c 5 0
2w2
5 16
x2
5 3x 3y 5 4 2 y2
7z 5 2z2
2 15
2w2
2 16 5 0 (no b-term)
x2
2 3x 5 0 (no c-term)
y2
1 3y 2 4 5 0 2z2
2 7z 2 15 5 0
Every quadratic equation has exactly two solutions, called roots (But the two roots might be the same.) On the GRE, you can often find the two roots by factoring To
solve any factorable quadratic equation, follow these three steps:
Put the equation into the standard form: ax2
1 bx 1 c 5 0.
Factor the terms on the left side of the equation into two linear expressions (with
no exponents)
Set each linear expression (root) equal to zero (0) and solve for the variable in each one
Factoring Simple Quadratic Expressions
Some quadratic expressions are easier to factor than others If either of the two
constants b or c is zero (0), factoring requires no sweat In fact, in some cases, no
factoring is needed at all:
A Quadratic with No c-term A Quadratic with No b-term
2x25 x 2x22 x 5 0
x~2x 2 1! 5 0
x 5 0, 2x 2 1 5 0
x 5 0,1
2
2x22 4 5 0
2~x22 2! 5 0
x22 2 5 0
x25 2
x 5=2, 2=2
Factoring Quadratic Trinomials
A trinomial is simply an algebraic expression that contains three terms If a quadratic
expression contains all three terms of the standard form ax2
1 bx 1 c, then factoring
becomes a bit trickier You need to apply the FOIL method, in which you add together these terms:
(F) the product of the first terms of the two binomials (O) the product of the outer terms of the two binomials (I) the product of the inner terms of the two binomials
(L) the product of the last (second) terms of the two binomials
ALERT!
When dealing with a
quadratic equation, your first
step is usually to put it into the
general form ax21 bx 1 c 5
0 But keep in mind: The only
essential term is ax2
Trang 3Note the following relationships:
(F) is the first term (ax2) of the quadratic expression
(O 1 I) is the second term (bx) of the quadratic expression
(L) is the third term (c) of the quadratic expression
You’ll find that the two factors will be two binomials The GRE might ask you to
recognize one or both of these binomial factors
25 Which of the following is a factor of x22 x 2 6 ?
(A) (x 1 6)
(B) (x 2 3)
(C) (x 1 1)
(D) (x 2 2)
(E) (x 1 3)
The correct answer is (B) Notice that x2 has no coefficient This makes the
process of factoring into two binomials easier Set up two binomial shells:
The product of the two missing second terms (the “L” term under the FOIL
method) is 26 The possible integral pairs that result in this product are (1,26),
(21,6), (2,23,), and (22,3) Notice that the second term in the trinomial is 2x.
This means that the sum of the two integers whose product is 26 must be 21
The pair (2,23) fits the bill Thus, the trinomial is equivalent to the product of the
two binomials (x 1 2) and (x 2 3).
To check your work, multiply the two binomials using the FOIL method:
~x 1 2!~x 2 3! 5 x22 3x 1 2x 2 6
5 x22 x 1 6
If the preceding question had asked you to determine the roots of the equation
x2
2 x 2 6 5 0, you’d simply set each of the binomial factors equal to zero (0), then solve
for x in each one The solution set (the two possible values of x) includes the roots 22
and 3
26 How many different values of x does the solution set for the equation
4x25 4x 2 1 contain?
(A) none
(B) one
(C) two
(D) four
(E) infinitely many
The correct answer is (B) First, express the equation in standard form: 4x22
Trang 4this product are (1,1) and (21,21) Since the b-term (24x) is negative, the
integral pair whose product is 1 must be (21,21) Set up a binomial shell: (? 2 1)(? 2 1)
Notice that the a-term contains the coefficient 4 The possible integral pairs that
result in this product are (1,4), (2,2), (21,24), and (22,22) A bit of trial-and-error reveals that only the pair (2,2) works Thus, in factored form, the equation
becomes (2x 2 1)(2x 2 1) 5 0 To check your work, multiply the two binomials
using the FOIL method:
~2x 2 1!~2x 2 1! 5 4x22 2x 2 2x 1 1
5 4x22 4x 1 1
Since the two binomial factors are the same, the two roots of the equation are the
same In other words, x has only one possible value (Although you don’t need to find the value of x in order to answer the question, solving for x in the equation 2x21 5 0 yields x 51
2.
Stealth Quadratic Equations
Some equations that appear linear (variables include no exponents) may actually be quadratic On the GRE, be on the lookout for either of two situations:
The same variable inside a radical also appears outside:
=x 5 5x
~=x!25 ~5x!2
x 5 25x2
25x22 x 5 0
The same variable that appears in the denominator of a fraction also appears elsewhere in the equation:
2
x 5 3 2 x
2 5 x~3 2 x!
2 5 3x 2 x2
x22 3x 1 2 5 0
In both scenarios, you’re dealing with a quadratic (nonlinear) equation with one variable So, in either equation, there are two roots (Both equations are factorable, so
go ahead and find their roots.) The test makers often use the Quantitative Comparison format to cover this concept
Trang 527. 6x= 3x
12
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (D) An unwary test taker might assume that the
equation is linear, because x is not squared Substituting 1
12 for x satisfies the
centered equation But the two quantities are not necessarily equal If you clear
the radical by squaring both sides of the equation, then isolate the x-terms on one
side of the equation, you’ll see that the equation is quadratic:
36 2 3
36 2 3 0
To find the two roots, factor out 3x, then solve for each root:
3x~12x 2 1! = 0
x = 0, 12x 2 1 = 0
x = 0, 1
12
Since 1
12 is only one of two possible values for x, you cannot make a definitive
comparison
NONLINEAR EQUATIONS WITH TWO VARIABLES
In the world of math, solving nonlinear equations with two or more variables can be
very complicated, even for bona-fide mathematicians But on the GRE, all you need to
remember are these three general forms:
Sum of two variables, squared: (x 1 y)25 x21 2xy 1 y2
Difference of two variables, squared: (x 2 y)25 x22 2xy 1 y2
Difference of two squares: x22 y25 (x 1 y)(x 2 y)
Trang 6You can verify these equations using the FOIL method:
(x 1 y)2
5 (x 1 y)(x 1 y)
5 x21 xy 1 xy 1 y2
5 x21 2xy 1 y2
(x 2 y)2
5 (x 2 y)(x 2 y)
5 x22 xy 2 xy 1 y2
5 x22 2xy 1 y2
(x 1 y)(x 2 y)
5 x21 xy 2 xy 2 y2
5 x22 y2
For the GRE, memorize the three equations listed here When you see one form on the exam, it’s a sure bet that your task is to rewrite it as the other form
28 If x2
2 y25 100, and if x 1 y 5 2, then x 2 y 5
(A) 22 (B) 10 (C) 20 (D) 50 (E) 200 The correct answer is (D) If you’re on the lookout for the difference of two
squares, you can handle this question with no sweat Use the third equation you
just learned, substituting 2 for (x 1 y), then solving for (x 2 y):
x22 y25 ~x 1 y!~x 2 y!
100 5 ~x 1 y!~x 2 y!
100 5 ~2!~x 2 y!
50 5 ~x 2 y!
SOLVING ALGEBRAIC INEQUALITIES
You can solve algebraic inequalities in the same manner as equations Isolate the variable on one side of the inequality symbol, factoring and eliminating terms wherever possible However, one important rule distinguishes inequalities from equa-tions: Whenever you multiply or divide both sides of an inequality by a negative
number, you must reverse the inequality symbol Simply put: If a b, then 2a , 2b.
12 2 4x , 8 Original inequality
24x , 24 12 subtracted from each side; inequality unchanged
x 1 Both sides divided by 24; inequality reversed Here are five general rules for dealing with algebraic inequalities Study them until they’re second nature to you because you’ll put them to good use on the GRE Adding or subtracting unequal quantities to (or from) equal quantities:
If a b, then c 1 a c 1 b
If a b, then c 2 a , c 2 b
TIP
You usually can’t solve
quadratics using a shortcut.
Always look for one of the
three common quadratic
forms If you see it, rewrite it as
its equivalent form to answer
the question as quickly and
easily as possible.
ALERT!
Be careful when handling
inequality problems: The wrong
answers might look right,
depending on the values you
Trang 7Comparing three unequal quantities:
If a b, and if b c, then a c
Combining the same positive quantity with unequal quantities by multiplication
or division:
If a b, and if x 0, then xa xb
Combining the same negative quantity with unequal quantities by multiplication
or division:
If a b, and if x , 0, then xa , xb
29 If a b, and if c d, then which of the following must be true?
(A) a 2 b c 2 d
(B) a 2 c b 2 d
(C) c 1 d , a 2 b
(D) b 1 d , a 1 c
(E) a 2 c , b 1 d
The correct answer is (D) Inequality questions can be a bit confusing, can’t
they? In this problem, you need to remember that if unequal quantities (c and d)
are added to unequal quantities of the same order (a and b), the result is an
inequality in the same order This rule is essentially what answer choice (D) says
SOLVING ALGEBRA “STORY” PROBLEMS
The remainder of this chapter is devoted exclusively to algebra “story” problems, in
which you translate real-world scenarios into algebraic expressions and equations
Here are the types of story problems you’ll review in the pages ahead (some call for the
application of specific formulas):
• weighted-average problems
• currency problems
• mixture problems
• investment problems
• problems involving rate of production and rate of travel
• problems involving overlapping sets
To illustrate each type, we’ll show you one or two GRE-style questions Read the
analysis of each question to find out what formulas to apply and what shortcuts (if
any) and strategies you can use to solve the problem
Weighted-Average Problems
You solve weighted-average problems using the arithmetic mean (simple average)
NOTE
For algebra story problems, the GRE test makers generally use the Problem Solving format (the same format we use here) Don’t be surprised, though, if your exam includes one or two Quantitative Comparisons that are actually story problems in disguise.
Trang 8think of the final exam score as two scores of 90—and the total number of scores as 4
rather than 3:
WA 575 1 85 1 ~2!~90!
340
4 5 85 Similarly, when some numbers among terms appear more often than others, you must give them the appropriate “weight” before computing an average
30 During an 8-hour trip, Brigitte drove 3 hours at 55 miles per hour and 5
hours at 65 miles per hour What was her average rate, in miles per hour, for the entire trip?
(A) 58.5 (B) 60 (C) 61.25 (D) 62.5 (E) 66.25 The correct answer is (C) Determine the total miles driven: (3)(55) 1 (5)(65) 5
490 To determine the average over the entire trip, divide this total by 8, which is the number of total hours: 490 4 8 5 61.25
A tougher weighted-average problem might provide the weighted average and ask for one of the terms, or require conversions from one unit of measurement to another—or both
31 A certain olive orchard produces 315 gallons of oil annually, on average,
during four consecutive years How many gallons of oil must the orchard produce annually, on average, during the next six years, if oil production for the entire 10-year period is to meet a goal of 378 gallons per year?
(A) 240 (B) 285 (C) 396 (D) 420 (E) 468 The correct answer is (D) In the weighted-average formula, 315 annual
gallons receives a weight of 4, while the average annual number of gallons for the
next six years (x) receives a weight of 6:
378 51,260 1 6x
10
3,780 5 1,260 1 6x 3,780 2 1,260 5 6x
2,520 5 6x
420 5 x
Trang 9To guard against calculation errors, check your answer by sizing up the
question Generally, how great a number are you looking for? Notice that the
stated goal is a bit greater than the annual average production over the first
four years So you’re looking for an answer that is greater than the goal—a
number somewhat greater than 378 gallons per year You can eliminate
choices (A) and (B) out of hand The number 420 fits the bill
Currency Problems
Currency problems are similar to weighted-average problems in that each item (bill or
coin) is weighted according to its monetary value Unlike weighted-average problems,
however, the “average” value of all the bills or coins is not at issue In solving currency
problems, remember the following:
• You must formulate algebraic expressions involving both number of items (bills or
coins) and value of items.
• You should convert the value of all moneys to a common currency unit before
formulating an equation If converting to cents, for example, you must multiply
the number of nickels by 5, dimes by 10, and so forth
32 Jim has $2.05 in dimes and quarters If he has four fewer dimes than
quarters, how much money does he have in dimes?
(A) 20 cents
(B) 30 cents
(C) 40 cents
(D) 50 cents
(E) 60 cents
The correct answer is (B) Letting x equal the number of dimes, x 1 4
represents the number of quarters The total value of the dimes (in cents) is 10x,
and the total value of the quarters (in cents) is 25(x 1 4) or 25x 1 100 Given that
Jim has $2.05, the following equation emerges:
10x 1 25x 1 100 5 205
35x 5 105
x 5 3
Jim has three dimes, so he has 30 cents in dimes
You could also solve this problem without formal algebra, by plugging in
each answer choice in turn Let’s try this strategy for choices (A) and (B):
(A) 20 cents is 2 dimes, so Jim has 6 quarters 20 cents plus $1.50 adds up
to $1.70 Wrong answer!
(B) 30 cents is 3 dimes, so Jim has 7 quarters 30 cents plus $1.75 adds up
TIP
You can solve most GRE currency problems by working backward from the answer choices.
Trang 10Mixture Problems
In GRE mixture problems, you combine substances with different characteristics, resulting in a particular mixture or proportion, usually expressed as percentages Substances are measured and mixed by either volume or weight—rather than by number (quantity)
33 How many quarts of pure alcohol must you add to 15 quarts of a solution
that is 40% alcohol to strengthen it to a solution that is 50% alcohol?
(A) 4.0 (B) 3.5 (C) 3.25 (D) 3.0 (E) 2.5 The correct answer is (D) You can solve this problem by working backward
from the answer choices—trying out each one in turn Or, you can solve the
problem algebraically The original amount of alcohol is 40% of 15 Letting x
equal the number of quarts of alcohol that you must add to achieve a 50% alcohol
solution, 4(15) 1 x equals the amount of alcohol in the solution after adding more alcohol You can express this amount as 50% of (15 1 x) Thus, you can express
the mixture algebraically as follows:
~0.4!~15! 1 x 5 ~0.5!~15 1 x!
6 1 x 5 7.5 1 0.5x 0.5x 5 1.5
x 5 3
You must add 3 quarts of alcohol to obtain a 50% alcohol solution
Investment Problems
GRE investment problems involve interest earned (at a certain percentage rate) on money over a certain time period (usually a year) To calculate interest earned, multiply the original amount of money by the interest rate:
amount of money 3 interest rate 5 amount of interest on money For example, if you deposit $1,000 in a savings account that earns 5% interest annually, the total amount in the account after one year will be $1,000 1 05($1,000)
5 $1,000 1 $50 5 $1,050
GRE investment questions usually involve more than simply calculating interest earned on a given principal amount at a given rate They usually call for you to set up and solve an algebraic equation When handling these problems, it’s best to eliminate percent signs