Simplifying Square Roots - ACT Math

Card 0 of 19

Question

Simplify:

$\sqrt[2]{24,300}$

Answer

$\sqrt[2]{24,300}$

When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.

What are the factors of 24,300?

24,300 can be factored into:

$24,300: 243\cdot 100$

When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is $10^{2}$ (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as: $10\sqrt[2]{243}$

But 243 can also be factored: $24,300: (3\cdot 81)\cdot100$
$24,300: {\color{Orange} 9}\cdot{\color{Orange} 9}\cdot {\color{Cyan} 10}\cdot {\color{cyan} 10}\cdot 3$
Following the same principle as for the 100, the problem would become
$9\cdot 10\sqrt[2]{3}$ because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9103.
9 and 10 may be multiplied together, yielding the final simplified answer of
$90\sqrt[2]{3}$

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Question

$\sqrt{180} + \sqrt{125} = ?$

Answer

To solve the equation $\sqrt{180} + \sqrt{125} = ?$ , we can first factor the numbers under the square roots.

$\sqrt{2\cdot 2\cdot 3\cdot 3\cdot 5} + \sqrt{5\cdot 5\cdot 5} = ?$

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

$2\cdot 3\sqrt{5} + 5\sqrt{5} = ?$

$6\sqrt{5} + 5\sqrt{5} = ?$

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

$6\sqrt{5} + 5\sqrt{5} = 11\sqrt{5}$

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Question

Which of the following is the most simplified form of:

$\sqrt{468}$

Answer

First find all of the prime factors of

$468=6\ast78=6\ast6\ast13=2\ast3\ast2\ast3\ast13$

So $\sqrt{468}=\sqrt{2\ast2\ast3\ast3\ast13}=2\ast3\sqrt{13}=6\sqrt{13}$

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Question

Which of the following is equal to $\sqrt{75}$ ?

Answer

√75 can be broken down to √25 * √3. Which simplifies to 5√3.

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Question

Simplify $\sqrt{a^{3}b^{4}c^{5}}$ .

Answer

Rewrite what is under the radical in terms of perfect squares:

$x^{2}=x\cdot x$

$x^{4}=x^{2}\cdot x^{2}$

$x^{6}=x^{3}\cdot x^{3}$

Therefore, $\sqrt{a^{3}b^{4}c^{5}}= \sqrt{a^{2}a^{1}b^{4}c^{4}c^{1}}=ab^{2}c^{2}\sqrt{ac}$ .

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Question

Which of the following is equivalent to $\frac{x + \sqrt{3}}{3x + \sqrt{2}}$ ?

Answer

Multiply by the conjugate and the use the formula for the difference of two squares:

$\frac{x + \sqrt{3}}{3x + \sqrt{2}}$

$\frac{x + \sqrt{3}}{3x + \sqrt{2}}\cdot \frac{3x - \sqrt{2}}{3x - \sqrt{2}}$

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{(3x)^{2} - (\sqrt{2})^{2}}$

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}$

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Question

What is $\sqrt{50}$ ?

Answer

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves $\sqrt{2}$ which can not be simplified further.

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Question

What is $\sqrt{432}$ equal to?

Answer

$\sqrt{432}$

1. We know that , which we can separate under the square root:

$\sqrt{144\cdot 3}$

2. 144 can be taken out since it is a perfect square: $12\cdot12=144$ . This leaves us with:

$12\sqrt{3}$

This cannot be simplified any further.

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Question

What is the simplified (reduced) form of $\sqrt{96}$ ?

Answer

To simplify a square root, you have to factor the number and look for pairs. Whenever there is a pair of factors (for example two twos), you pull one to the outside.

Thus when you factor 96 you get
$\ \sqrt{96}=\sqrt{2\cdot 2\cdot 2\cdot2\cdot2\cdot 3}\ \=2\cdot 2\sqrt{2\cdot 3}\ \=4\sqrt{6}$

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Question

Which of the following is equal to $\small \sqrt{1260}$ ?

Answer

When simplifying square roots, begin by prime factoring the number in question. For $\small 1260$ , this is:

$\small 1260=223*357$

Now, for each pair of numbers, you can remove that number from the square root. Thus, you can say:

$\small \sqrt{1260}=23\sqrt{35}=6\sqrt{35}$

Another way to think of this is to rewrite $\small \sqrt{1260}$ as $\small \sqrt{4}\sqrt{9}\sqrt{35}$ . This can be simplified in the same manner.

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Question

Which of the following is equal to $\small \sqrt{420}$ ?

Answer

When simplifying square roots, begin by prime factoring the number in question. For $\small 420$ , this is:

$\small 420 = 223*57$

Now, for each pair of numbers, you can remove that number from the square root. Thus, you can say:

$\small \small \sqrt{420}=2\sqrt{105}$

Another way to think of this is to rewrite $\small \small \sqrt{420}$ as $\small \sqrt{4}\sqrt{105}$ . This can be simplified in the same manner.

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Question

Which of the following is equivalent to $\small \sqrt{700700}$ ?

Answer

When simplifying square roots, begin by prime factoring the number in question. This is a bit harder for $\small 700700$ . Start by dividing out $\small 100$ :

$\small 700700 = 7007 * 100$

Now, $\small 7007$ is divisible by , so:

$\small \small 700700 =1001 * 7 * 2255$

$\small 1001$ is a little bit harder, but it is also divisible by $\small 7$ , so:

$\small 700700 =1437 7 2255$

With some careful testing, you will see that $\small 143 = 11 13$

Thus, we can say:

$\small \small 700700 = 22557 71113$

Now, for each pair of numbers, you can remove that number from the square root. Thus, you can say:

$\small \small \sqrt{700700} = 257\sqrt{143}=70\sqrt{143}$

Another way to think of this is to rewrite $\small \small \small \sqrt{700700}$ as $\small \sqrt{4} * \sqrt{25} * \sqrt{49} * \sqrt{143}$ . This can be simplified in the same manner.

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Question

Simplify the following square root:

$\sqrt{83}$

Answer

We need to factor the number in the square root and find pairs of factors inorder to simplify a square root.

Since 83 is prime, it cannot be factored.

Thus the square root is already simplified.

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Question

Right triangle $\Delta ABC$ has legs of length . What is the exact length of the hypotenuse?

Answer

If the triangle is a right triangle, then it follows the Pythagorean Theorem. Therefore:

--->

$c= \sqrt50$

At this point, factor out the greatest perfect square from our radical:

$\sqrt{50} = \sqrt{25 \cdot 2}$

Simplify the perfect square, then repeat the process if necessary.

$\sqrt{25 \cdot 2} = 5\sqrt2$

Since is a prime number, we are finished!

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Question

Simplify: $\sqrt{5000}$

Answer

There are two ways to solve this problem. If you happen to have it memorized that is the perfect square of , then $\sqrt{5000} = \sqrt{2500 \cdot 2} = 50\sqrt 2$ gives a fast solution.

If you haven't memorized perfect squares that high, a fairly fast method can still be achieved by following the rule that any integer that ends in is divisible by , a perfect square.

$\sqrt{5000} = \sqrt{25 \cdot 200} = 5\sqrt{200}$

Now, we can use this rule again:

$5\sqrt{200} = 5\sqrt{25 \cdot 8} = 25 \sqrt {8}$

Remember that we multiply numbers that are factored out of a radical.

The last step is fairly obvious, as there is only one choice:

$25 \sqrt{8} = 25\sqrt {4 \cdot 2} = 50\sqrt 2$

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Question

Simplify: $\sqrt{3240}$

Answer

A good method for simplifying square roots when you're not sure where to begin is to divide by , or , as one of these generally starts you on the right path. In this case, since our number ends in , let's divide by :

$\sqrt{3240} = \sqrt{324 \cdot 10}$

As it turns out, is a perfect square!

$\sqrt{324 \cdot 10} = 18\sqrt{10}$

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Question

Simplify: $8\sqrt{136}$

Answer

Again here, if no perfect square is easily recognized try dividing by , , or .

$8\sqrt{136} = 8\sqrt{4 \cdot 34} = 16 \sqrt{34}$

Note that the we obtained by simplifying $\sqrt{4}$ is multiplied , not added, to the already outside the radical.

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Question

Simplify:

$\sqrt{48}$

Answer

To solve, simply find a perfect square factor and pull it out of the square root.

Recall the factors of 48 include (16, 3). Also recall that 16 is a perfect square since 44=16.

Thus,

$\sqrt{48}\Rightarrow \sqrt{163}\Rightarrow \sqrt{4^2*3}\Rightarrow 4\sqrt{3}$

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Question

Solve:

$\sqrt{20}+\sqrt{45}=?$

Answer

The trick to these problems is to simplify the radical by using the following rule: $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $a\sqrt{b}+c\sqrt{b}=(a+c)\sqrt{b}$ Here, we need to find a common factor for the radical. This turns out to be five because $20=54\textup{ and }45=9*5.$ Remember, we want to include factors that are perfect squares, which are what nine and four are. Therefore, we can rewrite the equation as: $\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}.$

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