Radicals - College Algebra

Card 0 of 20

Question

Find the value of $\sqrt{80} + \sqrt{20}$ .

Answer

To solve this equation, we have to factor our radicals. We do this by finding numbers that multiply to give us the number within the radical.

$\sqrt{80} = \sqrt{20}\sqrt{4}$

$\sqrt{20} = \sqrt{5}\sqrt{4}$

Add them together:

$\sqrt5\sqrt4\sqrt4 + \sqrt5\sqrt4$

4 is a perfect square, so we can find the root:

$(2)(2)\sqrt5 + 2\sqrt5$

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

Since both have the same radical, we can combine them:

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

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Question

Multiply and express the answer in the simplest form:

$\small \sqrt3(\sqrt4-\sqrt3)$

Answer

$\small \sqrt3(\sqrt4-\sqrt3)$

$\small \sqrt{12}-\sqrt9$

$\small \sqrt{12}=\sqrt4 \times \sqrt3=2\sqrt3$

$\small \sqrt9=3$

$\small \sqrt{12}-\sqrt9=2\sqrt3-3$

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Question

Subtract and simplify:
$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Answer

$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Find the lease common denominator: $\small \sqrt{10}$

$\small \small \frac{\sqrt{3}}{\sqrt5}=\frac{\sqrt6}{\sqrt{10}}$

$\small \small \frac{\sqrt{3}}{\sqrt2}=\frac{\sqrt{15}}{\sqrt{10}}$

$\small \small \frac{\sqrt6}{\sqrt10}-\frac{\sqrt{15}}{\sqrt{10}}=\frac{\sqrt6-\sqrt{15}}{\sqrt{10}}$

A radical cannot be in the denominator:

$\small \small (\frac{\sqrt6-\sqrt{15}}{\sqrt{10}})(\frac{\sqrt{10}}{\sqrt{10}})=\frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \sqrt{60}=2\sqrt{15}$

$\small \sqrt{150}=5\sqrt6$

$\small \frac{\sqrt{60}-\sqrt{150}}{10}=\frac{2\sqrt{15}-5\sqrt6}{10}$

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Question

Simplify the following equation:

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

Answer

When simplifying, you should always be on the lookout for like terms. While it might not look like there are like terms in , there are -- we just have to be able to rewrite it to see.

$f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}$

$f(x)=x^{2}-3(4x)^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

Before we start combining terms, though, let's look a little more closely at this part:

$-3(4x)^{\frac{1}{2}}$

We need to "distribute" that exponent to everything in the parentheses, like so:

$-3\cdot 4^{\frac{1}{2}}\cdot x^{\frac{1}{2}}$

But 4 to the one-half power is just the square root of 4, or 2.

$-3\cdot 2\cdot x^{\frac{1}{2}}=-6x^{\frac{1}{2}}$

Okay, now let's see our equation.

$f(x)=x^{2}-6x^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}$

We need to start combining like terms. Take the terms that include x to the one-half power first.

$f(x)=x^{2}\mathbf{-6x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x\mathbf{-7x^{\frac{1}{2}}}+5x^{\frac{1}{3}}=x^{2}\mathbf{-13x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x+5x^{\frac{1}{3}}$

Now take the terms that have x to the one-third power.

$f(x)=x^{2}-13x^{\frac{1}{2}}\mathbf{+4x^{\frac{1}{3}}}-x\mathbf{+5x^{\frac{1}{3}}}=x^{2}-13x^{\frac{1}{2}}\mathbf{+9x^{\frac{1}{3}}}-x$

All that's left is to write them in order of descending exponents, then convert the fractional exponents into radicals (since that's what our answer choices look like).

$f(x)=x^{2}-x-13x^{\frac{1}{2}}+9x^{\frac{1}{3}}$

$f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}$

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Question

Simplify.

$\sqrt{36} \times 3\sqrt{36}$

Answer

We can solve this by simplifying the radicals first: $\sqrt{36} =6$

Plugging this into the equation gives us:

$6 \times3(6)=108$

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Question

What is the product of $\sqrt{12}$ and $3\sqrt{6}$ ?

Answer

First, simplify $\sqrt{12}=\sqrt{4\cdot3}$ to $2\sqrt{3}$ .

Then set up the multiplication problem:

$2\sqrt{3}\cdot 3\sqrt{6}$ .

Multiply the terms outside of the radical, then the terms under the radical:

$2\cdot 3\sqrt{3\cdot6}$ then simplfy: $6\sqrt{18}.$

The radical is still not in its simplest form and must be reduced further:

$6\sqrt{18}=6\cdot3\sqrt{2}=18\sqrt{2}$ . This is the radical in its simplest form.

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Question

Solve.

$\sqrt{3}+5\sqrt{3}+8\sqrt{3}$

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical. If they are not the same, the answer is just the problem stated.

Since they are the same, just add the numbers in front of the radical: which is

Therefore, our final answer is the sum of the integers and the radical:

$14\sqrt3$

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Question

Solve.

$\sqrt{5}+6\sqrt{5}-4\sqrt{5}$

Answer

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

If they are not the same, the answer is just the problem stated.

Since they are the same, just add and subtract the numbers in front: which is

Therefore, the final answer will be this sum and the radical added to the end:

$3\sqrt5$

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Question

Simplify, if possible: $\sqrt8+\sqrt{32}+\sqrt{50}$

Answer

The radicals given are not in like-terms. To simplify, take the common factors for each of the radicals and separate the radicals. A radical times itself will eliminate the square root sign.

$\sqrt8= \sqrt{4\times 2} = \sqrt4 \times \sqrt2 = 2\sqrt2$

$\sqrt{32} = \sqrt{4\times 4\times 2} = \sqrt4 \times \sqrt4 \times \sqrt 2 = 2\times 2\times \sqrt 2 = 4\sqrt2$

$\sqrt{50} = \sqrt{5\times 5 \times 2} = \sqrt5 \times \sqrt5 \times \sqrt2 = 5\sqrt2$

Now that each radical is in its like term, we can now combine like-terms.

$\sqrt8+\sqrt{32}+\sqrt{50} = 2\sqrt2 + 4\sqrt2 + 5\sqrt2 = 11\sqrt2$

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Question

Simplify the following:

$\sqrt[3]{x^6y^7z^8}$

Answer

To solve this, you must remember the rules for simplifying roots. In order to pull something out from the inside, you msut have the amount indicated in the index. Thus, in this case, to pull one x out, you need 3 inside. Thus,

$\sqrt[3]{x^6y^7z^8}=\sqrt[3]{(x^2)^3(y^2)^3(z^2)^3yz^2}=x^2y^2z^2\sqrt[3]{yz^2}$

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Question

Simplify the following:

$\sqrt{3}+\sqrt{5}+\sqrt{7}$

Answer

To simplify radicals, you must have common numbers on the inside of the square root. Don't be fooled. There is no way to simplify any of these, so your answer is simply:

$\sqrt{3}+\sqrt{5}+\sqrt{7}$

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Question

Simplify the following expression:

$\sqrt{250}+\sqrt{150}$

Answer

Observe that 250 and 150 factor into $25\cdot10$ and $25\cdot 6$ respectively. So,

$\sqrt{250}+\sqrt{150} = \sqrt{25\cdot 10}+\sqrt{25\cdot 6} = 5\sqrt{10}+5\sqrt{6}$

$=5(\sqrt{6}+\sqrt{10})$

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Question

Which of the following is equal to $5\sqrt{32}$ ?

Answer

Factor the radical by values of perfect squares.

$\sqrt{32}= \sqrt{16}\cdot \sqrt{2} = 4\sqrt2$

Replace the term.

$5\sqrt{32} = 5\cdot 4\sqrt2$

The answer is: $20\sqrt2$

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Question

Solve the radical: $\sqrt{3x+2} = 5$

Answer

Square both sides to eliminate the radical.

$(\sqrt{3x+2})^2 = 5^2$

Solve for x. Subtract two on both sides.

The answer is: $\frac{23}{3}$

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Question

What is the value of $\sqrt{3}\cdot \sqrt{15}\cdot \sqrt{2}$ ?

Answer

Multiply all numbers to combine the radicals.

$\sqrt{3}\cdot \sqrt{15}\cdot \sqrt{2} = \sqrt{90}$

Factor this value using numbers of perfect squares.

$\sqrt{90} = \sqrt{9\times 10} = \sqrt{9}\cdot \sqrt{10} = 3\sqrt{10}$

The answer is: $3\sqrt{10}$

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Question

Multiply the following radicals:

$\sqrt{5}\cdot \sqrt{20}$

Answer

In order to multiply radicals, first rewrite the problem as follows:

$\sqrt{5}\cdot \sqrt{20}\rightarrow \sqrt{5\cdot 20}$

This follows a basic property of radical multiplication.

Next simply inside the radical:

$\sqrt{5\cdot 20}\rightarrow \sqrt{100}$

Since 100 is a perfect square the final answer to the problem is 10.

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Question

Multiply the following radicals:

$\sqrt{2}\cdot \sqrt{10}$

Answer

In order to multiply radicals, first rewrite the problem as follows:

$\sqrt{2}\cdot \sqrt{10}\rightarrow \sqrt{2\cdot 10}$

This follows a basic property of radical multiplication.

Next simply inside the radical:

$\sqrt{2\cdot 10}\rightarrow \sqrt{20}$

Although 20 is not a perfect square, one of its factors is. We can break the radical up and simplify as follows:

$\sqrt{20}=\sqrt{4\cdot 5}=\sqrt{4}\cdot \sqrt{5}=2\cdot \sqrt{5}$

This gives us a final answer of $2\sqrt{5}$

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Question

Multiply the radicals:

$\sqrt{15}\cdot \sqrt{3}$

Answer

In order to multiply radicals, first rewrite the problem as follows:

$\sqrt{15}\cdot \sqrt{3}\rightarrow \sqrt{15\cdot 3}$

This follows a basic property of radical multiplication.

Next simply inside the radical:

$\sqrt{15\cdot 3}\rightarrow \sqrt{45}$

Although 45 is not a perfect square, one of its factors is. We can break the radical up and simplify as follows:

$\sqrt{45}=\sqrt{9\cdot 5}=\sqrt{9}\cdot \sqrt{5}=3\cdot \sqrt{5}$

This gives us a final answer of $3\sqrt{5}$

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Question

Multiply the radicals:

$\sqrt{2}\cdot \sqrt{6}$

Answer

In order to multiply radicals, first rewrite the problem as follows:

$\sqrt{2}\cdot \sqrt{6}\rightarrow \sqrt{2\cdot 6}$

This follows a basic property of radical multiplication.

Next simply inside the radical:

$\sqrt{2\cdot 6}\rightarrow \sqrt{12}$

Although 12 is not a perfect square, one of its factors is. We can break the radical up and simplify as follows:

$\sqrt{12}=\sqrt{4\cdot 3}=\sqrt{4}\cdot \sqrt{3}=2\cdot \sqrt{3}$

This gives us a final answer of $2\sqrt{3}$

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Question

Multiply the radicals:

$\sqrt{18}\cdot \sqrt{2}$

Answer

In order to multiply radicals, first rewrite the problem as follows:

$\sqrt{18}\cdot \sqrt{2}\rightarrow \sqrt{18\cdot 2}$

This follows a basic property of radical multiplication.

Next simply inside the radical:

$\sqrt{18\cdot 2}\rightarrow \sqrt{36}$

Since 36 is a perfect square the final answer to the problem is 6.

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