Radicals - Algebra II

Card 0 of 1760

Question

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

Answer

To solve this, remember that when multiplying variables, exponents are added. When raising a power to a power, exponents are multiplied. Thus:

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

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

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Question

Simplify by rationalizing the denominator:

$\frac{6}{\sqrt[3]{18} }$

Answer

Since $18 = 2 \cdot 3 ^{2}$ , we can multiply 18 by $2^{2} \cdot 3 = 12$ to yield the lowest possible perfect cube:

$18 \cdot 12 = 216 = 6 ^{3}$

Therefore, to rationalize the denominator, we multiply both nuerator and denominator by $\sqrt[3]{12}$ as follows:

$\frac{6}{\sqrt[3]{18} }$

$= \frac{6 \cdot \sqrt[3]{12} }{\sqrt[3]{18} \cdot \sqrt[3]{12} }$

$= \frac{6 \cdot \sqrt[3]{12} }{\sqrt[3]{18 \cdot 12 } }$

$= \frac{6 \sqrt[3]{12} }{\sqrt[3]{216 } }$

$= \frac{6 \sqrt[3]{12} }{6 }$

$= \sqrt[3]{12}$

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Question

Rationalize the denominator and simplify:

$\frac{1}{6+\sqrt{27}}$

Answer

To rationalize a denominator, multiply all terms by the conjugate. In this case, the denominator is $6+\sqrt{27}$ , so its conjugate will be $6-\sqrt{27}$ .

So we multiply: $\frac{1}{6+\sqrt{27}} \times \frac{6-\sqrt{27}}{6-\sqrt{27}} = \frac{6-\sqrt{27}}{36-27}$ .

After simplifying, we get $\frac{6-\sqrt{27}}{36-27} = \frac{3\cdot2-3\sqrt{3}}{9}=\frac{2-\sqrt{3}}{3}$ .

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Question

Simplify: $\sqrt[3]{120000}$

Answer

Begin by getting a prime factor form of the contents of your root.

Applying some exponent rules makes this even faster:

Put this back into your problem:

Returning to your radical, this gives us:

$\sqrt[3]{2^635^4}$

Now, we can factor out sets of and set of . This gives us:

$2^25\sqrt[3]{35}=20\sqrt[3]{15}$

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Question

Simplify:

$\sqrt[4]{15625}$

Answer

Begin by factoring the contents of the radical:

This gives you:

$\sqrt[4]{5^6}$

You can take out group of . That gives you:

$5\sqrt[4]{5^2}=5\sqrt[4]{25}$

Using fractional exponents, we can rewrite this:

$55^\frac{2}{4}$

Thus, we can reduce it to:

$55^\frac{1}{2}$

Or:

$5\sqrt{5}$

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Question

Simplify: $\sqrt{125}+\sqrt{50}$

Answer

To simplify $\sqrt{125}+\sqrt{50}$ , find the common factors of both radicals.

$\sqrt{125}= \sqrt{25\cdot5}=\sqrt{25}\cdot \sqrt5=5\sqrt5$

$\sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot \sqrt2=5\sqrt2$

Sum the two radicals.

The answer is: $5\sqrt5+5\sqrt2$

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Question

Simplify:

$\sqrt[3]{27x^5y^6}$

Answer

To take the cube root of the term on the inside of the radical, it is best to start by factoring the inside:

$\sqrt[3]{x^3\cdot x^2\cdot y^6\cdot 27}$

Now, we can identify three terms on the inside that are cubes:

We simply take the cube root of these terms and bring them outside of the radical, leaving what cannot be cubed on the inside of the radical.

$x\cdot y^2 \cdot 3 \cdot \sqrt[3]{x^2}$

Rewritten, this becomes

$3xy^2\sqrt[3]{x^2}$

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Question

Simplify the radical: $\sqrt{32}\cdot\sqrt{8}$

Answer

Simplify both radicals by rewriting each of them using common factors.

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

$\sqrt{8} = \sqrt4 \cdot \sqrt2 = 2\sqrt2$

Multiply the two radicals.

$4\sqrt2\cdot 2\sqrt2 = 4\cdot 2 \cdot 2 = 16$

The answer is:

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Question

Simplify: $-\sqrt{ 120}$

Answer

In order to simplify this radical, rewrite the radical using common factors.

$-\sqrt{ 120} = -\sqrt{2} \cdot \sqrt{2}\cdot \sqrt{5} \cdot \sqrt{6}$

Simplify the square roots.

$-2\cdot \sqrt{5} \cdot \sqrt{6}$

Multiply the terms inside the radical.

The answer is: $-2\sqrt{30}$

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Question

Simplify: $\sqrt{30} \cdot \sqrt{15}$

Answer

Break down the two radicals by their factors.

$\sqrt{30} \cdot \sqrt{15} = (\sqrt{3}\times\sqrt{2}\times \sqrt{5})\cdot(\sqrt{3}\times \sqrt{5})$

A square root of a number that is multiplied by itself is equal to the number inside the radical.

$\sqrt{3}\cdot \sqrt{3}=3$

$\sqrt{5}\cdot \sqrt{5}=5$

Simplify the terms in the parentheses.

$(\sqrt{3}\times\sqrt{2}\times \sqrt{5})\cdot(\sqrt{3}\times \sqrt{5}) = 3\times 5 \times \sqrt2$

The answer is: $15\sqrt2$

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Question

Simplify, if possible: $3\sqrt{5}+6\sqrt{50}+ 9\sqrt{500}$

Answer

The first term is already simplified. The second and third term will need to be simplified.

Write the common factors of the second radical and simplify.

$6\sqrt{50} =6\sqrt{25\times 2} = 6\sqrt{25}\times \sqrt{2} = 6(5)\sqrt2 = 30\sqrt2$

Repeat the process for the third term.

$9\sqrt{500} = 9\sqrt{100\times 5} = 9\sqrt{100}\times \sqrt{5} = 9(10)\sqrt{5} = 90\sqrt5$

Rewrite the expression.

$3\sqrt{5}+6\sqrt{50}+ 9\sqrt{500}= 3\sqrt{5}+30\sqrt2+90\sqrt5$

Combine like-terms.

The answer is: $30\sqrt2+93\sqrt5$

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Question

What is the value of $\sqrt{288}+\sqrt{48}$ ?

Answer

Simplify the first term by using common factors of a perfect square.

$\sqrt{288} = \sqrt{144\times 2} = \sqrt{144} \cdot \sqrt{2} = 12\sqrt2$

Simplify the second term also by common factors.

$\sqrt{48} =\sqrt{4\cdot 4\cdot 3} =\sqrt{4}\cdot \sqrt{4}\cdot \sqrt{3} = 4\sqrt3$

Combine the terms.

$\sqrt{288}+\sqrt{48}=12\sqrt2+4\sqrt3$

The coefficients cannot be combined since these are unlike terms.

The answer is: $12\sqrt2+4\sqrt3$

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Question

Simplify: $\frac{\sqrt{32}}{\sqrt{6}}$

Answer

To simplify this, multiply the top and bottom by the denominator.

$\frac{\sqrt{32}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt2\cdot \sqrt6}{6} =\frac{4\sqrt{12}}{6} =\frac{4\cdot 2\sqrt{3}}{6}$

Reduce the fraction.

The answer is: $\frac{4\sqrt3}{3}$

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Question

Simplify: $\sqrt{927}$

Answer

To simplify radicals, we need to find perfect squares to factor out. In this case, it's .

$\sqrt{927}=\sqrt{9}*\sqrt{103}=3\sqrt{103}$ .

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Question

Simplify: $\sqrt{1248}$

Answer

To simplify radicals, we need to find perfect squares to factor out. In this case, it's .

$\sqrt{1248}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$

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Question

Simplify: $\sqrt{x^7y^3}$

Answer

To simplify radicals, we need to find perfect squares to factor out. In this case, it's .

$\sqrt{x^7y^3}=\sqrt{x^6}\sqrt{y^2}\sqrt{xy}=\sqrt{(x^3)^2}*y\sqrt{xy}=x^3y\sqrt{xy}$

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Question

Simplify: $\sqrt{128x^4y^5}$

Answer

To simplify radicals, we need to find perfect squares to factor out. In this case, it's .

$\sqrt{128x^4y^5}=\sqrt{64}\sqrt{(x^2)^2}\sqrt{(y^2)^2}*\sqrt{2y}=8x^2y^2\sqrt{2y}$

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Question

Simplify: $2\sqrt{60} \cdot 3\sqrt{10}$

Answer

Multiply the radicals.

$\sqrt{60} \times \sqrt{10} = \sqrt{600}$

Simplify this by writing the factors using perfect squares.

$\sqrt{600} = \sqrt{100} \times \sqrt{6} = 10\sqrt6$

Multiply this with the integers.

$2\sqrt{60} \cdot 3\sqrt{10} = 2\cdot 3 \cdot 10\sqrt6 = 60\sqrt6$

The answer is: $60\sqrt6$

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Question

Simplify: $10\sqrt{20}\cdot 30\sqrt{40}$

Answer

Multiply the integers and combine the radicals together by multiplication.

$10\sqrt{20}\cdot 30\sqrt{40} =300 \sqrt{800}$

Break up square root of 800 by common factors of perfect squares.

$300 \sqrt{800} = 300 \times \sqrt{100}\times \sqrt{4}\times \sqrt2$

Simplify the possible radicals.

$300\times 10 \times 2 \times \sqrt2 = 6000\sqrt2$

The answer is: $6000\sqrt2$

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Question

Evaluate: $5\sqrt{30}\cdot 4\sqrt{10}$

Answer

Multiply the integers and the value of the square roots to combine as one radical.

$5\sqrt{30}\cdot 4\sqrt{10} = 20\sqrt{300}$

Simplify the radical. Use factors of perfect squares to simplify root 300.

$20\sqrt{300} = 20 \sqrt{100} \cdot \sqrt{3}= 20\cdot 10\cdot \sqrt3$

The answer is: $200\sqrt3$

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