First, a few practice problems.

1) The numbers a, b, and c are all positive.  If , then what is the value of ?

Statement #1: a – b = 3

Statement #2: = 7

2) Given that (P + 2Q) is a positive number, what is the value of (P + 2Q)?

Statement #1: Q = 2

Statement #2:

3) In the diagram above, O is the center of the circle, DC = a and DO = b.  What is the area of the circle?

Statement #1:

Statement #2: a + b = 22

4) ABCD is a square with a side y, and JKLM is a side x.  If Rectangle S (not shown) with length (x + y) has the same area as the shaded region above, what is the width of Rectangle S?

(A) x

(B) y

(C) y + x

(D) y – x

(E)

Three important algebra patterns

Doing math involve both following procedures and recognizing patterns.  Three important patterns for algebra on the GMAT are as follows:

Pattern #1: The Difference of Two Squares

Pattern #2: The Squares of a Sum

Pattern #3: The Squares of a Difference

For GMAT Quant success, you need to know these patterns cold.  You need to know them as well as you know your own phone number or address.  The GMAT will throw question after question at you in which you simply will be expected to recognize these patterns.  In such a question, if you recognize the relevant formula, it will enormously simplify the problem.  If you don’t recognize the relevant formula, you are likely to be stymied by such a question.

Memory, not memorizing

You might think I would say: memorize them.  Instead, I will ask you to remember them.  What’s the difference?  Memorization implies a rote process, simply trying to stuff an isolated and disconnected factoid into your head.  By contrast, you strengthen you capacity to remember a math formula when you understand all the logic that underlies it.

Here, the logic behind these formulas is the logic of FOILing and factoring.  You should review those patterns until you can follow each both ways — until you can FOIL the product out, or factor it back into components.  If you can do that, you really understand these, and are much more likely to remember them in an integrated way.

Summary

If these patterns are relatively new to you, you may want to revisit the problems at the top with the list handy: see if you can reason your way through them, before reading the explanations below.  Here’s another practice problem from inside Magoosh:

5) http://gmat.magoosh.com/questions/129

Do you have questions?  Is there anything you would like to say?  Let us know in the comment section at the bottom!

Practice Problem Explanations

1) Let X = , the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get = X – 117.  In other words, if we could find the value of , then we could find the value of X.

Statement #1: a – b = 3

From this statement alone, we cannot calculate , so we can’t find the value of X.  Statement #1, alone and by itself, is insufficient.

Statement #2: = 7

From this statement alone, we cannot calculate , so we can’t find the value of X.  Statement #2, alone and by itself, is insufficient.

Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21.  Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out , and if we know the numerical value of that, we can calculate X and answer the prompt.   Combined, the statements are sufficient.

Answer = C

2) The prompt tells us that (P + 2Q) is a positive number, and we want to know the value of P.  Remember number properties! We don’t know that (P + 2Q) is a positive integer, just a positive number of some kind.

Statement #1: Q = 2

Obvious, by itself, this tells us zilch about P.  Alone and by itself, this statement is completely insufficient.

Statement #2:

Now, this may be a pattern-recognition stretch for some folks, but this is simply the “Square of a Sum” pattern.  It may be clearer if we re-write it like this:

This is now the “Square of a Sum” pattern, with P in the role of A and 2Q in the role of B.  Of course, this should equal the square of the sum:

All we have to do is take a square root.  Normally, we would have to consider both the positive and the negative square root, but since the prompt guarantees that (P + 2Q) is a positive number, we need only consider the positive root:

This statement allows us to determine the unique value of (P + 2Q), so this statement, alone and by itself, is sufficient.

Answer = B

3) To find the area of the circle, we need to use Archimedes’ formula, . For that we need the radius, OC.  We are not given this directly, but notice that r = OC = DC – OD = a – b.  If we knew that, we could find the area of the circle.

Statement #1:

A major pattern-matching hit!  This, as written, is the “Square of a Difference” pattern.

In fact, this statement already gives us , so we just have to multiply by pi and we have the area.  This statement, alone and by itself, is sufficient.

Statement #2: a + b = 22

We need a – b, and this statement gives us a value of  a + b.  If we had more information, perhaps we could use this in combination with other information to find what we want, but since this is all we have, it’s simply not enough to find a – b.  This statement, alone and by itself, is insufficient.

Answer = A

4) A tricky one.  First of all, notice that the shaded area, quite literally and visually, is the difference of two square — Area = .  We know from the Difference of Two Squares pattern, this factors into:

Area = = (y + x)(y – x)

Well, if a rectangle had this same area, and it had a length of (y + x), it would have to have to have a width of (y – x) — that would make the area the same.  The width has to be (y – x).  Answer = D

The post Three Algebra Formulas Essential for the GMAT appeared first on Magoosh GMAT Blog.

This is a potentially tricky topic.  First, consider these practice questions.

1. Which value(s) of x satisfies the equation above?

I. –1

II. 4

III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

2. What is the value of y?

A full discussion of these problems will come at the end of this article.

A subtle distinction

Consider the following, very simple algebra equation

Of course, this equation has two solutions, +5 and -5.  Of course, we find that by taking the square-root of both sides, remembering this process involves a ± sign.  Now, by contrast, consider this expression:

Many folks might think this is, in all respects, the same as the solution in the first process — i.e. “taking the square root.”  We need to draw a subtle distinction here, concerning the nature of this sign:

The benighted unfortunately will refer to this as a “square root” sign, but that is a misleading partial name.  The name of this simple is the “principal square root” sign, where the word “principal”, in the sense of “main” or “primary”, here means the positive root only.  Accordingly, the output of this sign is always positive.

The number 25 has two square roots, one positive and one negative, but it has only one principal square root.  A number can have, at most, only one principal square root.  The principal square root of 25 is +5 only.

When a squared algebraic expression appears in a problem, and we ourselves, in the process of problem-solving, find a square root, we need to include all roots, positive and negative (a common mistake is to forget the negative roots).  BUT, when the symbol above, the principle square root symbol, is printed on the page as part and parcel of the given problem, this means its output will always be positive.

Thus, the paradoxical juxtaposition

Solving an equation with a radical

By “equation with a radical”, I will be referring to any algebraic equation in which some of the algebra is under a radical sign, a.k.a a principal square root sign.  The general strategy for such an equation is (a) isolate the radical (i.e. get it alone by itself on one side of the equation); (b) square both sides, thus eliminating the radical; and (c) solve what remains using algebra (often, this will involve factoring a quadratic.)  All well and good, but there’s a catch.

You see, when you square both sides of equation, sometimes that creates solutions that weren’t part of the original equation.  Consider the hyper-simple equation x = 5.  This “equation” has only one solution, positive five.  BUT, if we square both sides, then we get the equation we solved in the previous section, with solutions ±5.  The extra root, x = -5, was not a solution of the original equation, but it became a solution once we squared.  This is an example of an extraneous root —- a number that is not a root of the original equation, but which “becomes” a root when we square both sides.

We have to square both sides to solve an equation with radicals, but doing so introduces the possibility of an extraneous root.  Thus, an essential part of solving any equation with radicals is to check the answers you find, in order to ascertain whether any are extraneous roots.  We verify the roots by plugging them into the original equation —- if the number does not solve the original equation, as given in the problem, then it is not a bonafide solution.

BTW, just as a general point of strategy, regardless of whether radicals are involved, I recommend checking any algebra values you find by plugging them back into the original equation given in the problem, when possible.  It’s just a good habit to check your work.

Extraneous roots play a role in both of the sample problems above.  Having read this post, you may want to go back and give them another attempt before reading the solutions below.  Questions?  Let us know in the comment section at the bottom.

Solutions to the sample questions

1) The radical is already isolated, so square both sides.

(x – 9)(x + 1) = 0

preliminary solutions: x = {+9, –1}

At this point, an unsuspecting student might be tempted to answer (C), the trap answer.  BUT, the problem is: one or both of these answers could be extraneous.  We need to check each by plugging back into the original expression.

Check x = –1

This answer does not check — the left & right sides have different values.  Thus, x = –1 is an extraneous root, not a solution to the problem.

(NB: it’s often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Check x = 9

The value x = 9 checks — it makes the two sides equal, and thus satisfies the original equation.  This is the only solution, so only option III contains a root.

Answer = (B)

2) Using a tried and true DS strategy, start with the easier statement, Statement #2.

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

The radical is already isolated, so square both sides.

Lo and behold!  We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2.  The naïve conclusion would be — this statement says exactly the same thing as the other.  That’s incorrect, though, because we don’t know whether both of these values are valid solutions, or whether one or more is an extraneous root.  We need to test this in the original equation.

Test y = +4

This value checks — y = +4 is a valid solution to the equation

Test y = –2

The two sides are not equal, so this does not check!  This value, y = –2, is an extraneous root.

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question.  This statement, alone and by itself, is sufficient.

Answer = (A)

The post GMAT Math: Algebra Equations with Radicals appeared first on Magoosh GMAT Blog.

The GMAT sometimes features quant questions with strange symbols. These symbols should not fluster you too much as long as you remember that they do not represent standard mathematical notation. Instead, the symbols pertain only to the problem and are defined by the GMAT (or whatever prep material you happen to be using).

Let’s have a look at a simple example:

. What is the value of ?

(A) 2
(B) 3
(C) 5
(D) 6
(E) 9

Explanation

To approach strange symbols think of the Q’ as a recipe. To the right of the equals sign are the steps (or the recipe) you have to follow.

Another way of looking at it, whatever we see in place of Q’ we want to plug it into the ‘Q’ in . Therefore . Because the question has two apostrophe signs, we want to repeat this procedure to get, . Answer (D).

This is a basic problem, one that if you saw it on the GMAT, would not bode well. So let’s try a problem that will make you sweat a little more.

. What the value of p in ?
(A) -5
(B) 9
(C) 13
(D) 25
(E) 625

Explanation

Be careful not to fall the trap that switches the order of b and a. Our equation should read: . Solving for p:

.

Answer (E).

For those who are looking to score a Q51, here are two brutally difficult questions. If you think you know the answer, go ahead and post it below with an explanation.

Brutal Question #1

. What is the difference between the least and the greatest possible values of x + y, if x@y is an integer less than 15?

(A) 9
(B) 10
(C) 49
(D) 50
(E) 52

Brutal Question #2

[[x]] is equal to the lesser of the two integer values closest to non-integer x. What is the absolute value of ?

(A)

(B)

(C)

(D)

(E)

Answers:

Brutal Question #1 – E
Brutal Question #2 – E

The post GMAT Math: Strange Symbols appeared first on Magoosh GMAT Blog.

First, here are two challenging Quant problems, both involving three equations with three algebraic unknowns.

2a + b + 3c = 6

a – b + 5c = 12

3a + 2b + 2c = 2

1) Given the equations above, what does the product a*b*c equal?

(A) 6
(B) -6
(C) 12
(D) -12
(E) 24

6x – 5y + 3z = 23

4x + 8y – 11z = 7

5x – 6y + 2z = 12

2) Given the equations above, x + y + z = ?

(A) 11
(B) 12
(C) 13
(D) 14
(E) 15

Full solutions will follow this article.

Simplifying

In a previous post, I discussed solving two algebra equations with two unknowns.  That’s already a challenging task.  Three equations with three unknowns is even trickier, and something you are quite unlikely to see unless you are already performing brilliantly on the Quant section.

Well, it turns out, solving these, we use a time-honored problem-solving strategy: inside every big problem is a little problem struggling to get out. Yes, that’s playfully stated, but I have found it is often surprisingly apt in all kinds of personal and interpersonal situations.

What’s considerably more pertinent here — this basic idea is the core of much advanced mathematical thinking, at the levels of calculus, analysis, number theory, and other more abstruse topics. At all levels, mathematicians strive to reduce problems which they don’t know how to solve to problems which they do know how to solve.  That can be a hugely valuable perspective on GMAT mathematical strategy.  Among other things, that’s precisely the approach with these problems. I will assume you have read that previous post and are somewhat competent in the two variable/two equation problems — for the purposes of this discussion, I will consider those the problems we do know how to solve. I will also assume you are familiar with substitution and elimination from that previous post.

The approach

Here’s the general strategy for solving three equations with three unknowns.

Step #1: Pick a pair of equations, two of the three, and using either substitution or elimination, eliminate one of the variables.  Most often, elimination is much much easier than substitution!  After this step, we will end up with one equation with two unknowns.

For this one, you have to step back and have your right-brain pattern matching hat on.  You have to think very strategically about what would be the most efficient.  For example, in problem #1 above, if I wanted to pick the first two equations and eliminate c, I could do that, but it would involve multiplying the first equation by 5 and the second equation by (-3), which would lead to some big numbers.  Hmmm.  Not the slickest approach.  Instead, when I look at these equations I notice —- the first equation has a (+b) and the second equation has (-b), so without any fuss, I could add those equations and right away eliminate b.  That’s a considerably more efficient approach.

Step #2: Pick a different pair of equations, and through elimination, eliminate the same variable you eliminated in step #1.  As the result of this step, we now will have two equations with the same two unknowns.

Step #3: At this point, we have reduced the problem we didn’t know how to solve to one we do know how to solve: two equations with two unknowns.  Use those techniques to solve for those two variables.

Step #4: Once you have the numerical values of two of the variable, plug into any of the original equations to solve for the value of the third variable.

I will demonstrate this entire strategy in the solution to #1 below.

Do I have to solve?

For problem #1 above, we have to solve fully, but the coefficients are reasonably small, and as it turns out, the numbers come out nice and neat.  By contrast, #2 is a monster.  The numbers are larger and uglier, and the answer will come up as ugly fractions.   But, as it turns out, we can answer the questions being asked with only a minimum of calculations.

This is where you really have to have your creative, out-of-the-box, right brain cap on.   Question #2 is not asking for the values of individual variables, but for an expression.  As it turns out, there’s an unbelievably simple way to jump directly to the answer with astonishingly little work.  Do you see it?  I will discuss this in the solution below.

The Moral: Don’t automatically assume you always have to slog through the hard work of solving for all the individual variables.   Always keep your antennae up for creative, time-saving shortcuts!

Summary

Once again, these problems are very rare.   You will not see them at all unless you are performing at 700+ level, getting almost everything else right.   Here’s another problem, for further practice:

3) http://gmat.magoosh.com/questions/1009

If you have any questions on what I’ve said here, let me know in the comments sections below!

Solutions to practice problems

1) Here, I will show the full solution outlined above.  First of all, here are the equations, with letter designations.

(P) 2a + b + 3c = 6

(Q) a – b + 5c = 12

(R) 3a + 2b + 2c = 2

(I started later in the alphabet, so these letter-names of the equations wouldn’t be confused with answer choice letters!)  First of all, I notice that lovely (+b) in (P) and (-b) in (Q), so I will add those two.

That is new equation (S), with variable a & c.  Now, in step #2, we want to pick a different pair of equations, and eliminate the same variable, b.  Again, I like the (-b) in (Q) — that won’t be hard to use to cancel (+2b) in equation in (R).  Just multiply (Q) by 2, and add it to (R).

Now, with (S) and (T), we have two equations with the same two unknowns, a & c.  The numbers 8 & 24, the coefficients of c, have a LCM of 24.  Multiply (S) by 3 and (T) by (-2).

Thus, a  = -2.  Plug this value into (S).

Plug these two values into (P).

Thus, {a, b, c} = (-2, 1, 3), and their product a*b*c = -6.  Answer = (B).

2) As with the last problem, I will begin by giving letter names to the equations.

(P) 6x – 5y + 3z = 23

(Q) 4x + 8y – 11z = 7

(R) 5x – 6y + 2z = 12

Here, it would be a colossal waste of time to solve for the individual values of x & y & z separately.  We want to find the value of x + y + z.  Notice, first of all, that the x-coefficient of (R) is one higher than that of (Q); unfortunately, their y-coefficients are 14 units apart from each other, not close at all.  Now, notice that the x-coefficient of (P) is one higher than that of (R); also, the y-coefficient of (P) is one higher than that of (R); also, the z-coefficient of (P) is one higher than that of (R)!  BINGO!  The difference, (P) – (R), equals the expression we seek!

Answer = A

The post GMAT Quant: Three Equations, Three Unknowns appeared first on Magoosh GMAT Blog.

More formally, these are called “rational expressions” in mathematics — “rational” in the sense of “having do with a ratio”, as the word is used in the phrase “rational numbers“.   These rational expression appear in some of the most challenging Quant problems on the GMAT.   Here are a few practice questions.

(A) –2
(B) –1
(C) 0
(D) 1
(E) 2

(A) zero
(B) one
(C) two
(D) a finite number greater than two
(E) infinitely many

5. As y increases from y = 247 to y = 248, which of the following decreases?

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Solutions will follow this article.

Thoughts on rational algebraic expression

First of all, to understand this stuff, you should be clear on the basic rules of fractions: how to add, subtract, multiply, and divide them.  If you can’t do this basic arithmetic with numerical fractions, it will be very hard to do it with algebraic rational expressions!   Some further tips:

1.  Suppose you have an equation involving one or more algebraic rational expressions.  Suppose you are asked to solve for values of the variable.  It’s important to note that any value of the variable that makes any individual denominator equal to zero cannot possibly be a solution of the equation.   This can be a powerful tool in “which of the following could be a solution” question, because usually you can immediately eliminate a few answers right away, which sets you up very well for backsolving or solution behavior.

2. When adding or subtracting rational expressions, as when adding or subtracting ordinary fractions, we must find a common denominator to combine.  We must do precisely the same thing with rational expressions.  Here are a couple examples of this process.

Example #1

Example #2

3.  Whenever you have just one algebraic fraction on one side of the equation equal to just one algebraic fraction on the other side of the equation, then you can cross-multiply. If either side has more than one fraction, added or subtracted, you would have to combine them, via the previous hint, before you are ready to cross-multiply.

4.  If the whole equation has only one or two denominators, you also can simply multiply every term on both sides by the denominators.  That can be a very efficient way to get rid of all the fractions in one fell swoop.  For example, the equation:

can be simplified by multiplying each term by (x – 2) —- with the fraction, it cancels the denominator, simply leaving the numerator.

x(x – 2) = 2(x – 2) + 1

If we were to multiply this all out, we would get a quadratic that we could solve.

5.  One could always use a direct algebraic solution: that may be efficient or that may take several steps and be time-consuming, even if you know you are doing. Remember that backsolving may be quicker. For a compound fraction (a big fraction with a little fraction in the numerator or denominator), it may well be quicker to step back and perform a more holistic solution, looking at what must be true about each piece: I demonstrate this in the solution for practice problem #2 below.

Summary

If this article gave you any insights, you may want to give the practice problems another look before jumping into the explanations below.  Here’s another practice question from inside Magoosh.

6. http://gmat.magoosh.com/questions/137

If you would like to express anything or ask for clarification, please let us know in the comments section below.

Explanations to the practice problems

1) We could multiply by the three denominators, (x – 2), then (x – 1), then (x + 1), get three quadratics, and simplify.  That’s a lot of work.

We could find a common denominator on the right, combine them into a single ugly fraction, then when we had a single fraction on each side, we could cross multiply.  That’s also a lot of work.

The simple solution for which this problem is crying out is backsolving.  Notice that three of the answer choices — (A) & (D) & (E) — are illegal because each makes one of the denominators zero.  We can immediately eliminate all three of those.  Now, it’s just a matter of plugging in the other two answer choices.

The two sides are not equal, so (B) can’t be the right answer.   At this point, we pretty much know that (C) must be the answer, but it’s always good to verify that it works.

Both sides are equal, so x = 0 satisfies this equation.  Therefore, answer = (C).

2) Rather than do a ton of algebraic re-arranging, let’s think about this.  We have 3 divided by (something) equals 1/2.  This means, the “something” must equal 6.   That immediately produces the much simpler equation:

Answer = (A)

3) Multiply all three terms by x and we get

This equation is unfactorable.  It is not a perfect square.  Think about its graph, which is a parabola:

When x = 0, y is negative, and when x = 2, y is positive.  Therefore, the parabola intersects the x-axis twice, which means the equation has two real solutions.

Answer = (C).

BTW, this is a special mathematical equation.  One solution is the Golden Ratio, and the other solution is the negative reciprocal of the Golden Ratio.

4) This is an easy one to solve.  Subtract 2 from both sides:

Now, add 3/y to both sides.  Because the two fractions have the same denominator, y, we can just add the numerators:

Answer = (E)

5) As y gets larger, what happens to each one of these?

For statement I, as y gets larger, the 2y gets larger.  Since the subtracting 100 stays the same as the value of y changes, that makes no difference.  This one increases as y increases, so it is not a correct choice.

For statement II, as the denominator of a fraction increases, the value of the fraction overall decreases.   When y increases, 50/y has to decrease.  Again, adding 80 remains the same as y changes, so this doesn’t make any difference.  This is a correct choice.

For statement III, as long as y > 3, then y^2 – 3y will increase as y increase.  That means the entire fraction decrease.  We are subtracting 100 minus the fraction, and if the fraction gets smaller, then we are subtracting something smaller and therefore are left with more.   This means the entire expression, the difference, gets bigger as y increase.   This one increases as y increases, so it is not a correct choice.

Answer = (B)

The post Algebraic Fractions on the GMAT appeared first on Magoosh GMAT Blog.

A lot of students prepping for GMAT Quant, especially those GMAT students away from math for a long time, get lost when trying to divide by a square root. However, dividing by square roots is not something that should intimidate you. With a short refresher course, you’ll be able to divide by square roots in no time.

Practice questions: How to divide by a square root

First, consider these three practice questions.

1.  In the equation above, x =

2. Triangle ABC is an equilateral triangle with an altitude of 6.  What is its area?

3.  In the equation above, x =

The second one throws in a little geometry.  You may want to review the properties of the 30-60-90 Triangle and the Equilateral Triangle if those are unfamiliar.  The first one is just straightforward arithmetic.  The third is quite hard.  For any of these, it may well be that, even if you did all your multiplication and division correctly, you wound up with an answers of the form —something divided by the square root of something—and you are left wondering: why doesn’t this answer even appear among the answer choices?  If this has you befuddled, you have found exactly the right post.

Fractions and radicals

When we first met fractions, in our tender prepubescence, both the numerators and denominators were nice easy positive integers.  As we now understand, any kind of real number, any number on the entire number line, can appear in the numerator or denominator of a fraction.  Among other things, radicals—that is, square-root expressions—can appear in either the numerator or denominator.  There’s no particular issue if we have the square-root in a numerator.  For example,

is a perfectly good fraction.  In fact, those of you who ever took trigonometry might even recognize this special fraction.   Suppose, though, we have a square root in the denominator: what then?  Let’s take the reciprocal of this fraction.

This is no longer a perfectly good fraction.  Mathematically, this is a fraction “in poor taste”, because we are dividing by a square root.  This fraction is crying out for some kind of simplification.  How do we simplify this?

Dealing with square roots in the denominator

By standard mathematical convention, a convention the GMAT follows, we don’t leave square-roots in the denominator of a fraction.  If a square-root appears in the denominator of a fraction, we follow a procedure called rationalizing the denominator.

We know that any square root times itself equals a positive integer.  Thus, if we multiplied a denominator of the square root of 3 by itself, it would be 3, no longer a radical.  The trouble is—we can’t go around multiplying the denominator of fractions by something, leaving the numerator alone, and expect the fraction to maintain its value.  BUT, remember the time-honored fraction trick—we can always multiply a fraction by A/A, by something over itself, because the new fraction would equal 1, and multiplying by 1 does not change the value of anything.

Thus, to simplify a fraction with the square root of 3 in the denominator, we multiply by the square root of 3 over the square root of 3!

That last expression is numerically equal to the first expression, but unlike the first, it is now in mathematical “good taste”, because there’s no square root in the denominator.  The denominator has been rationalized (that is to say, the fraction is now a rational number).

Sometimes, some canceling occurs between the number in the original numerator and the whole number that results from rationalizing the denominator.  Consider the following example:

That pattern of canceling in the simplification process may give you some insight into practice problem #1 above.

Square roots and addition in the denominator

This is the next level of complexity when it comes to dividing by square roots.  Suppose we are dividing a number by an expression that involves adding or subtracting a square root.  For example, consider this fraction:

This is a fraction in need of rationalization.  BUT, if we just multiply the denominator by itself, that WILL NOT eliminate the square root — rather, it will simply create a more complicated expression involving a square root.  Instead, we use the difference of two squares formula, = (a + b)(a – b).  Factors of the form (a + b) and (a – b) are called conjugates of one another.  When we have (number + square root) in the denominator, we create the conjugate of the denominator by changing the addition sign to a subtraction sign, and then multiply both the numerator and the denominator by the conjugate of the denominator.   In the example above, the denominator is three minus the square root of two.  The conjugate of the denominator would be three plus the square root of two.  In order to rationalize the denominator, we multiply both the numerator and denominator by this conjugate.

Notice that the multiplication in the denominator resulted in a “differences of two squares” simplification that cleared the square roots from the denominator.  That final term is a fully rationalized and fully simplified version of the original.

Summary

Having read these posts about dividing by square roots, you may want to give the three practice questions at the top of this article another try, before reading the explanations below.  If you have any questions on dividing by square roots or the explanations below, please ask them in the comments sections! And good luck conquering these during your GMAT!

Practice question explanations

1) To solve for x, we will begin by cross-multiplying. Notice that

because, in general, we can multiply and divide through radicals.

Cross-multiplying, we get

You may well have found this and wondered why it’s not listed as an answer.  This is numerically equal to the correct answer, but of course, as this post explains, this form is not rationalized.  We need to rationalize the denominator.

Answer = (D)

2) We know the height of ABC and we need to find the base.  Well, altitude BD divides triangle ABC into two 30-60-90 triangles.  From the proportions in a 30-60-90 triangle, we know:

Now, my predilection would be to rationalize the denominator right away.

Now, AB is simplified. We know AB = AC, because the ABC is equilateral, so we have our base.

Answer = (C)

3) We start by dividing by the expression in parentheses to isolate x.

Of course, this form does not appear among the answer choices.  Again, we need to rationalize the denominator, and this case is a little trickier because we have addition in the denominator along with the square root.  Here we need to find the conjugate of the denominator—changing the plus sign to a minus sign—and then multiply the numerator and denominator by this conjugate.  This will result in:

Answer = (A)

The post GMAT Math: How to Divide by a Square Root appeared first on Magoosh GMAT Blog.