Basic Squaring / Square Roots - ACT Math

To solve the equation , we can first factor the numbers under the square roots.

When a factor appears twice, we can take it out of the square root.

Now the numbers can be added directly because the expressions under the square roots match.

First, we can simplify the radicals by factoring.

Now, we can factor out the .

Now divide and simplify.

To begin with, factor out the contents of the radicals. This will make answering much easier:

They both have a common factor . This means that you could rewrite your equation like this:

This is the same as:

These have a common . Therefore, factor that out:

These three roots all have a in common; therefore, you can rewrite them:

Now, this could be rewritten:

Now, note that

Therefore, you can simplify again:

Now, that looks messy! Still, if you look carefully, you see that all of your factors have ; therefore, factor that out:

This is the same as:

Begin by factoring out the relevant squared data:

is the same as

This can be simplified to:

Since your various factors contain square roots of , you can simplify:

Technically, you can factor out a :

The most difficult element of this problem is being able to correctly simplify each of the square roots. This means starting with our first contestant, . There are several different methods for simplifying square roots, each with their own benefits. Furthermore, some calculators permissible on the ACT can actually do it for you (might be worth looking into). But for those of us not so fortunate, here is one particular method, which utilizes what is called a factor tree.

Start by writing the number inside the square root, 45 in our case. And find two factors of (numbers that multiplied together equal) 45 other than 1 and the number itself. In our case there are two possibilities: 9 and 5 or 3 and 15. Both will work, but we will go with the latter pair. Look at each of the two factors to see if the same process can be repeated. Looking at 3, we realize it is prime, meaning no additional factors exist. Therefore, we leave the 3 as is. However, the 15 can be further broken up into the factors 5 and 3, which we can illustrate as shown. We then attempt to repeat the process with the next tier of numbers, but we soon realize that both 5 and 3 are prime. Therefore, we can go no further, and our factor tree is finished.

But what do we do now? We look for pairs. For each pair of numbers we have under the radical, we put one number on the outside of the square root. That means one 3 goes outside the square root. For each under the radical without a pair, we put that number inside the square root, meaning one 5 goes inside the square root. This gives a final answer of .

We then repeat the process with our second contestant, . A factor tree for 180 could take several different paths, but any correct one (including the example below) should end with the same numbers under the radical. Though perhaps not in the same order, the numbers under the radical should include two 2s, two 3s, and one 5. We see that a pair of 2s and a pair of 3s should give us one 2 and one 3 on the outside of the square root, while a pairless 5 should go on the inside. Any time multiple numbers end up on either the inside or outside, we simply multiple those numbers.

Therefore, we get .

We then complete the process one more time with our final contestant, 125. Doing so invariably gives the following factor tree.

The problem in this case is that we don't have a pair but a trio of 5s. In this case, we simply pair two of them, leaving one to be the odd five out.

Our final answer for this contestant then is .

Substituting our simplified square roots for the originals gives us the new expression

. From here it is best to think of like an apple. In the first term I have 3 "apples". I then subtract, or take away, 6 "apples". Finally, I add back 5 "apples". How many apples do I have?

. I have 2 apples, or in other words .

Begin by breaking apart the square roots on the left side of the equation:

This can be rewritten:

You can combine like terms on the left side:

Solve by dividing both sides by :

This simplifies to:

To begin solving this problem, find the greatest common perfect square for all quantities under a radical.

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Pull out of each term on the left:

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Next, factor out from the left-hand side:

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Lastly, isolate :

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To begin solving this problem, find the greatest common perfect square for all quantities under a radical.

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Factor out the square root of each perfect square:

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Next, factor out from each term on the left-hand side of the equation:

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Lastly, isolate :

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Solving this one is tricky. At first glance, we have no common perfect square to work with. But since each term can produce the quantity , let's start there:

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Simplify the first term:

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Divide all terms by to simplify,

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Next, factor out from the left-hand side:

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Isolate by dividing by and simplifying:

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Last, simplify the denominator:

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Right away, we notice that is a prime radical, so no simplification is possible. Note, however, that both other radicals are divisible by .

Our first step then becomes simplifying the equation by dividing everything by :

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Next, factor out from the left-hand side:

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Lastly, isolate :

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Once again, there are no common perfect squares under the radical, but with some simplification, the equation can still be solved for :

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Simplify:

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Factor out from the left-hand side:

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Lastly, isolate :

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To begin solving this problem, find the greatest perfect square for all quantities under a radical. might seem intimidating, but remember that raising even single-digit numbers to the fourth power creates huge numbers. In this case, is divisible by , a perfect fourth power.

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Pull the perfect terms out of each term on the left:

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Next, factor out from the left-hand side:

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Lastly, isolate , remembering to simplify the fraction where possible:

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To start, begin pulling the largest perfect square you can out of each number:

So, . You can just add the two terms together once they have a common radical.

Again here, it is easiest to recognize that both of our terms are divisible by , a prime number likely to appear in our final answer:

Now, simplify our perfect squares:

Lastly, subtract our terms with a common radical:

Take each fraction separately first:

(2√3)/(√2) = \[(2√3)/(√2)\] * \[(√2)/(√2)\] = (2 * √3 * √2)/(√2 * √2) = (2 * √6)/2 = √6

Similarly:

(4√2)/(√3) = \[(4√2)/(√3)\] * \[(√3)/(√3)\] = (4√6)/3 = (4/3)√6

Now, add them together:

√6 + (4/3)√6 = (3/3)√6 + (4/3)√6 = (7/3)√6

Find the Sum:

Simplify the radicals:

The first step when adding square roots is to simplify each term as much as possible. Since the first term has a square within the square root, we can reduce to . Now each term has only prime numbers within the sqaure roots, so nothing can be further simplified and our new expression is .

Only terms that have the same expression within the sqaure root can be combined. In this question, these are the first and third terms. When combining terms, the expression within the square root stays the say, while the terms out front are added. Therefore we get . Since the second term of the original equation cannot be combined with any other term, we get the final answer of .

Begin by factoring out each of the radicals:

For the first two radicals, you can factor out a or :

The other root values cannot be simply broken down. Now, combine the factors with :

This is your simplest form.

Begin by getting your terms onto the left side of the equation and your numeric values onto the right side of the equation:

Next, you can combine your radicals. You do this merely by subtracting their respective coefficients:

Now, square both sides:

Solve by dividing both sides by :