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)**

I’ve NEVER done this kind of math in my life and am still trying to wrap my mind around the square root of 56 divided by 3, but I enjoyed how well you explained the equations

Why did you not use the area of an equilateral to solve this problem? What is the difference between this problem and the problem posted in the link below?

https://gmat.magoosh.com/answers/153510727?&review%5Bafter%5D=2017-04-20+00%3A00%3A00+-0500

Thank You

Hi Chinel,

You can definitely use the formula of the area of an equilateral to solve this problem: (√3/4)a². However, there are definitely other approaches, and you don’t need to memorize this formula. But, if you’d like to, you can. You still first need to solve for the side of the equilateral (“a”) first.

Now, with regards to the second question, you’re right they are very similar. The only difference is the first step of calculating/deriving the side of the equilateral triangle! Once you’ve got this, you’re golden! 😀

Hello, My question is similar to Suru’s for problem 2. How do we know AB/BD = 2/root 3?

I know that 30-60-90 triangles have ratio of sides of x, root 3 , and 2. But BD has value 6.

Was something canceled to get 2/root 3?

Hi Anish,

AB/BD is just the ratio of the hypotenuse/side of a 30-60-90 triangle. Mike set up this equation to determine the relationship between the two sides, so that he can determine the value of AB. The relationship between these two sides of a 30-60-90 triangle will always be 2/root3, which he then sets equal to the values given in the practice problem. We know BD but not AB, so he creates the expression AB/6=2/root3 in order to determine the value of AB. Does that make sense? You can read more about ratios and how to set them up here 🙂

I’m trying to solve a practice question in regard to hypothesis testing. The question is 0.3889-0.5 / 0.5 * (1-0.5) / 36 although there is a square root on the bottom of the equation is square root over 0.5 * (1-0.5) / 36. The answer comes to -1.3332 however I can’t reach the answer. Please help me.

Hi Zina,

First of all, you’re not likely to receive a question like this on the actual GMAT (given the numerator). If anything, this would be an approximation/estimation problem, and the answer choices would need to reflect that. Still, let’s break this problem down. Let’s first write this problem with all the information you provided:

= (0.3889-0.5) / (0.5*(1-0.5)/36)^(1/2)

Next, let’s work on the numerator first:

= Numerator

= (0.3889-0.5)

= (0.3889-0.5000)

= -0.1111

Then, let’s work on the denominator:

= Denominator

= (0.5*(1-0.5)/36)^(1/2)

= ((1/2)*(1/2)/36)^(1/2)

= ((1/4)/36)^(1/2)

= (1/(4*36))^(1/2)

= (1/4)^(1/2) * (1/36)^(1/2)

= (1/2) * (1/6)

= 1/12

Now, let’s combine the two:

= Numerator/Denominator

= -0.1111/(1/12)

= -0.1111 * 12

= -1.3332

I hope this helps a little!

what would this equation equal in terms of x. I’m trying to solve this as a practice question, but its come up as really difficult for me.

Square root of H^2 = square root of (x^2 – 1/4x^2)

I’m trying to turn this into an equation along the lines of H = Nx

Thank You very much

Happy to help! 🙂 Right away, you can turn “Square root of H^2 = square root of (x^2 – 1/4x^2)” into just H = x – 1/4X. This is because the square root of x^2— or of any number ^2 is just the original number. From there, you just need to simplify x – 1/4x. Expressing this as 1x – 1/3x, you can easily see that the simplification is 3/4x. To put this in steps:

STEP 1: Square root of H^2 = square root of (x^2 – 1/4x^2)

STEP 2: H = x – 1/4x

STEP 3: H = 3/4x

So in your H = Nx term, N would be 3/4.

Hope this helps. 🙂 Let me know if you have any questions, Jayden.

Just saw this. For your explanation to be accurate, doesn’t there have to be a multiplication in the parenthesis, instead of subtraction or addition?

Hi Philip,

I’m happy to clarify but I’m not exactly sure which step you’re referring to. If you could specify what part of the explanation seems inaccurate, I’ll definitely take another look to see if something is off! Thanks 🙂

Hi Mike,

Thanks for this post. I completely understand everything on this page- thanks for that- except for the 30-60-90 triangle yielding AB/BD=2/3^0.5

I feel like this is something I should know, but I still can’t figure it out. Can you please explain a bit, or point me to what lesson covers this. I’m sure I’ll come across it soon, but I wont have peace of mind until I understand it now 🙂

Thank you!

Toby.

Hi Toby,

I’ve sent this to our team of experts since you’re a premium student. You should hear from them soon. 🙂

How can I solve this problem is root 2/2and both are cancelled send me the answer in = root.

Hi Suru 🙂

I’m happy to help but I’m not sure which of the practice problems you’re referring to. Let us know and we’ll get back to you as soon as we can 🙂

Thanks!

Hi i do not know how to divide 13 by the square root by 3… can you help

Hi Mya,

Sorry for the late reply! I’m happy to help. 🙂

We are going to follow the procedure described in the post and in this image:

If we have 13/√3 we need to multiply the top and bottom of the fraction by √3 to eliminate the root on the bottom. This leads to (13*√3)/(√3*√3). Once we do the multiplication, that simplifies to (13√3)/3! I hope that helps! 🙂

Hi,

your write that “that pattern of canceling in the simplification process may give you some insight into practice problem #1 above”.

But then, all we needed to do is cross mulitply. So to me it seems like that pattern was not used

Lukas,

I’m happy to respond. 🙂 Yes, the first step was to cross-multiply, but one we did that and solved for x, we got 24/sqrt(6), and as I pointed out, that is not listed as one of the answer choices.

That’swhen we needed the pattern of canceling that I discussed—afterthe cross multiplication.Does this make sense?

Mike 🙂

For the first question, how come when you cross multiply it isn’t:

24= sqrt(2) * (x * sqrt(3))

I thought when you cross-multiply you have to multiply by the whole quantity not just select terms?

Dear Malcolm,

I’m happy to respond. 🙂 When we cross multiply, it IS 24 = sqrt(2)*(x*sqrt(3)), but because sqrt(2)*sqrt(3) = sqrt(6), as I pointed out at the beginning of that explanation, I just skipped a step, and jumped right to x*sqrt(6).

Much in the same way, if we had x/5 = 5/3, cross-multiplying gives 3x = 25 — we don’t have to write out the 5*5, because it’s obvious that’s 25. In much the same way, multiplication of radicals needs to be obvious to you.

Does all this make sense?

Mike 🙂

Nevermind – I completely forgot to do the parenthesis first (I was thinking x * sqrt(2) + sqrt(6)). But after a closer look I see what you are saying.

Thanks!

Dear Malcolm,

You are more than welcome, my friend. Best of luck to you!

Mike 🙂

Hello, thanks for explaining these… very helpful.

One question on the equilateral triangle though. I followed up until we calculate the side as 4root3. Being an equilateral triangle, shouldn’t we calculate the area as:

Root 3 / 4 ( side^2)

or

Root 3 / 4 (4 root 3)^2 ?

Dear Wasiff,

I’m happy to respond. 🙂 First of all, let’s be clear.

Fact #1: for every single triangle possible, every single triangle conceivable, the formula A = (0.5)b*h will give the area. There is not a single triangle in creation whose area cannot be found by that formula. This is the general universal rule for triangle area.Fact #2: for some individual special triangles, three are, in addition to the general universal area formula, also special case area formulas that work.Fact #3: for an equilateral triangle, the formula A = (3/4)(s^2) is the optional “shortcut” formula that also works in addition to the general universal formula. In some cases, it’s easier to use this shortcut for an equilateral triangle. In some cases, the universal formula is actually easier to use for equilateral triangle than is the shortcut.This problem here is in the latter case — a problem in which, yes, we

coulduse the specialized “shortcut” rule, but the “shortcut” would actually be a bit more complicated here (because the side contains a radical). It’s actually quicker and easier in this problem to use the generalized rule.Does all this make sense?

Mike 🙂

Makes sense. Thank you for explaining that!

Dear Wasiff,

You are more than welcome. Best of luck to you!

Mike 🙂

Hi Mike,

For question 2.

Triangle ABC is an equilateral triangle with an altitude of 6. What is its area?as the equilateral triangle is cut in half wouldn’t that make it a 45,45,90 triangle as 90/2 = 45 ? therefore side BC would be 6 root 2 and the area would be root 3/ 4 * (side bc squared) 6 root 2 squared ? 6 root 2 squared would be 36*2 = 72 so 72* root 3/4 which would be 18 root 3

Dear Cody:

I’m happy to respond. 🙂 An equilateral triangle has three 60 degree angles. When we cut an equilateral triangle down the middle, we get a 30-60-90 triangle, NOT a 45-45-90 triangle. (When we cut a square along a diagonal, we get a 45-45-90 triangle.)

Does this make sense?

Mike 🙂