Exponents And Roots On The GMAT

by on September 15th, 2009

Exponents and roots are common in the quantitative section of the GMAT for one main reason: most test takers are not adept at manipulating them (multiplying, dividing, factoring, taking roots, and so on). Let’s take a look at some shortcuts to help.

Finding and estimating square roots

Without using a calculator, find the square root of the perfect square 576.

Most people will use a trial-and-error approach, squaring a bunch of numbers until they hit upon the one that works. A savvier approach is to use benchmarks to zero in on the answer.

  • 20^2 is 400, and
  • 25^2 is 625

So the square root of 576 is somewhere between 20 and 25.

Because 576 is a perfect square, the square root is an integer, so the square root is either 21, 22, 23, or 24. The units digit of 576 is 6, so the units digit of its square root must yield a 6 when squared. The units digits 1, 2, and 3 do not yield a 6 when squared; only 4 yields a 6 when squared. The square root must therefore be 24.

What if the number is not a perfect square? We can still use this process to estimate. Let’s try 873. This lies between:

  • 25^2 (625), and
  • 30^2 (900).

In fact, 873 is so close to 900 that its root must be very close to 30. We can estimate its root to be approximately 29. Check it on a calculator: the square root is approximately 29.5.

Finding fourth roots

All even exponents are squares. Obviously, x^2 is a square, but so is x^76. How do we know?

Because x^76 = (x^38)(x^38).

The square root of any even exponent, then, is simply the base raised to half the exponent. For example:

sqrt{x^52} = x^26, since (x^26)(x^26) = x^52

All we are doing is dividing the exponent by 2. Elaborating on this principle, we can find the fourth root (the square root of the square root) of x^52 by dividing the exponent by 4:

root{4}{x^52} = x^13 because x^52 = (x^13)(x^13)(x^13)(x^13)

Okay, suppose we’re asked this question:

If x^12 = 2, what is the value of x^6?

Well, if x^12 = 2, then

x^6 = sqrt{x^12} = sqrt{2}

Therefore, x^ 6 = sqrt{2}

The difference of two squares

What is the “difference of two squares” anyway? Simply put, it is one square minus another:

x^2 - y^2

Maybe you have vague recollections of this expression from high-school algebra. If so, perhaps you recall that it can be factored in a predictable pattern:

x^2 - y^2 right (x + y)(x - y)

Basically all we are doing is multiplying the sum of the square roots by the difference of the square roots. It is this pattern that interests the GMAT. On a basic level, the GMAT wants to see that we know how to factor the difference of two squares given in a straightforward form such as

x^2 - y^2

As the exam gets harder, however, we need to be a little more supple in our handling of the difference of two squares.

First, we need to recognize that expressions like x^22 -  y^18 can be treated as the difference of two squares, because all even exponents are also squares:

x^22 - y^18 right (x^11 + y^9)(x^11 - y^9)

We can simplify an expression such as:

{x^22- y^18}/{x^11+y^9}

by factoring the numerator (top) as the difference of two squares:

{x^22-y^18}/{x^11+y^9} right {(x^11+y^9)(x^11-y^9)}/{x^11+y^9} right x^11-y^9

Second, we need to recognize that even expressions like x – y can be factored as the difference of two squares:

x - y right (sqrt{x} + sqrt{y})(sqrt{x} - sqrt{y})

Again, all we are doing is multiplying the sum of the square roots by the difference of the square roots. We can use this knowledge to help us simplify expressions such as

{x - y}/{sqrt{x}-sqrt{y}}

by factoring the numerator as the difference of two squares:

{x - y}/{sqrt{x} - sqrt{y}} right {(sqrt{x} + sqrt{y})(sqrt{x} - sqrt{y})}/{sqrt{x} - sqrt{y}} right sqrt{x} + sqrt{y}

Third, we need to recognize that we can use the factored form of the difference of two squares to help simplify expressions such as

3/{sqrt{6}+sqrt{5}} (because the GMAT frowns upon radicals in denominators)

Since the denominator contains the sum of two radicals, we can multiply top and bottom by the complement


to rid the denominator of the radicals:

{3/{sqrt{6}+sqrt{5}}}({sqrt{6}-sqrt{5}}/{sqrt{6}-sqrt{5}}) right {3(sqrt{6}-sqrt{5})}/{6-5} right 3sqrt{6}-3sqrt{5}

Exponents and consecutive integers

Let’s say we have an integer x. If we see (x)(x + 1), we know that this is the product of 2 consecutive integers: x and the next integer up. But if we want to be slippery, we can express this relationship as:

x^2 + x

All we have done is multiply out (x)(x+1), but now it is not as obvious that we are looking at the product of two consecutive integers.

If we multiply x by the next integer down, we get (x – 1)(x). To make the consecutive integer designation less obvious, we can distribute the x:

x^2 - x

If we continue the pattern with (x – 1)(x)(x + 1), we have three consecutive integers. But, again, if we want to be slippery, we can distribute:

(x^2 - x)(x + 1) = x^3 + x^2 - x^2 - x = x^3 - x

Now, x^3 - x is considerably less recognizable as the product of three consecutive integers.

What use does this have? On the GMAT, these relationships are often used to test our understanding of the concept of odd and even. How?

Consider x^2 - x, where x is an integer.

This expression must be even. First, we can look at it as an odd integer multiplied by an even integer (remember, all pairs of consecutive integers consist of an even number and an odd number), which will always yield an even integer. Or we can look at x itself:

If x is even, then x^2 - x is even minus even, which is even. If x is odd, then x^2 - x is odd minus odd, which is even.

These relationships come up much more often in Data Sufficiency than in Problem Solving, for the simple reason that they convey information in a sneaky way. The GMAT uses them as shorthand to see whether you understand the code. When you practice, if you see a particular expression that was used to hide a consecutive relationship, write that down and memorize it as an alternative form. For example, we saw above that

(x - 1)(x)(x + 1) = x^3 - x

If we’re given:

x^3 - x (and we are told x is an integer)

then we also know that we have three consecutive integers!

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