Multiplying And Dividing Radical Expressions

Multiplying Radical Expressions

When multiplying radical expressions with the same index, nosotros use the production rule for radicals. Given existent numbers $\sqrt[n]{A}$ and $\sqrt[n]{B}$ ,

$$\sqrt[n]{A}\cdot \sqrt[\mathrm{due\; north}]{B}=\sqrt[\mathrm{due\; north}]{A\cdot B}$$

Example one

Multiply: $\sqrt[3]{12}\cdot \sqrt[3]{6}.$

Solution:

Employ the product rule for radicals, then simplify.

$$\begin{array}{lllll}\hfill \sqrt[\mathrm{iii}]{12}\cdot \sqrt[3]{6}& =\hfill & \sqrt[3]{12\cdot \mathrm{six}}\hfill & \hfill & \mathrm{Thousand}ultiplytheradicand\mathrm{south}.\hfill \\ \hfill & =\hfill & \sqrt[3]{72}\hfill & \hfill & Simplify.\hfill \\ \hfill & =\hfill & \sqrt[3]{2^{\mathrm{iii}}\cdot {\mathrm{three}}^2}\hfill & \hfill & \hfill \\ \hfill & =\hfill & 2\sqrt[3]{{\mathrm{three}}^2}\hfill & \hfill & \hfill \\ \hfill & =\hfill & 2\sqrt[\mathrm{iii}]{9}\hfill & \hfill & \hfill \end{array}$$

Reply: $2\sqrt[\mathrm{iii}]{9}$

Oftentimes, in that location volition be coefficients in front of the radicals.

Example 2

Multiply: $\mathrm{iii}\sqrt{6}\cdot \mathrm{five}\sqrt{\mathrm{ii}}$

Solution:

Using the product dominion for radicals and the fact that multiplication is commutative, we tin multiply the coefficients and the radicands every bit follows.

$$\begin{array}{llll}3\sqrt{6}\cdot 5\sqrt{2}\hfill & =\hfill & \mathrm{iii}\cdot 5\cdot \sqrt{6}\cdot \sqrt{\mathrm{two}}\hfill & Multip\mathrm{fifty}icationisco\mathrm{thousand}kutati5e\text{.}\hfill \\ \hfill & =\hfill & 15\cdot \sqrt{12}\hfill & \mathrm{Grand}u\mathrm{50}tiplyth\mathrm{east}co\mathrm{due\; east}fficientsand\hfill \\ \hfill & \hfill & \hfill & theradicand\mathrm{southward}\text{.}\hfill \\ \hfill & =\hfill & 15\sqrt{4\cdot \mathrm{three}}\hfill & Si\mathrm{1000}plify\text{.}\hfill \\ \hfill & =\hfill & 15\cdot 2\cdot \sqrt{3}\hfill & \hfill \\ \hfill & =\hfill & 30\sqrt{3}\hfill & \hfill \end{array}$$

Typically, the first stride involving the awarding of the commutative belongings is not shown.

Reply: $\mathrm{xxx}\sqrt{3}$

Case three

Multiply: $-3\sqrt[3]{4y^2}\cdot 5\sqrt[\mathrm{three}]{\mathrm{xvi}y}.$

Solution:

$$\begin{array}{lllll}\hfill \text{−}3\sqrt[3]{\mathrm{iv}{y}^{\mathrm{ii}}}\cdot 5\sqrt[3]{16y}& =\hfill & \text{−}\mathrm{fifteen}\sqrt[3]{64y^3}\hfill & \hfill & \mathrm{Thou}ultiplythecoefficie\mathrm{due\; north}t\mathrm{southward}a\mathrm{northward}dthe\mathrm{northward}\mathrm{yard}ultiplytheradicands.\hfill \\ \hfill & =\hfill & \text{−}\mathrm{xv}\sqrt[3]{{\mathrm{four}}^3{y}^3}\hfill & \hfill & \mathrm{Due\; south}i\mathrm{grand}p\mathrm{fifty}ify.\hfill \\ \hfill & =\hfill & \text{−}\mathrm{fifteen}\cdot 4y\hfill & \hfill & \hfill \\ \hfill & =\hfill & \text{−}60y\hfill & \hfill & \hfill \end{array}$$

Answer: $\text{−}60y$

Use the distributive property when multiplying rational expressions with more than one term.

Example 4

Multiply: $5\sqrt{2\mathrm{ten}}\left(\mathrm{iii}\sqrt{x}-\sqrt{\mathrm{ii}x}\right)$ .

Solution:

Apply the distributive property and multiply each term by $5\sqrt{2x}.$

$$\begin{array}{lllll}\hfill \mathrm{five}\sqrt{2x}\left(3\sqrt{x}-\sqrt{\mathrm{two}\mathrm{10}}\right)& =\hfill & \mathrm{v}\sqrt{\mathrm{two}\mathrm{10}}\cdot \mathrm{iii}\sqrt{x}-\mathrm{five}\sqrt{\mathrm{two}x}\cdot \sqrt{2\mathrm{10}}\hfill & \hfill & Distribut\mathrm{east}.\hfill \\ \hfill & =\hfill & 15\sqrt{2x^2}-5\sqrt{4x^2}\hfill & \hfill & Si\mathrm{one\; thousand}p\mathrm{50}ify.\hfill \\ \hfill & =\hfill & 15x\sqrt{2}-\mathrm{five}\cdot 2x\hfill & \hfill & \hfill \\ \hfill & =\hfill & \mathrm{fifteen}x\sqrt{2}-10x\hfill & \hfill & \hfill \end{array}$$

Respond: $15x\sqrt{2}-10x$

Example 5

Multiply: $\sqrt[3]{6x^{2}y}\left(\sqrt[3]{\mathrm{ix}{\mathrm{ten}}^2{y}^2}-\mathrm{five}\cdot \sqrt[3]{\mathrm{four}\mathrm{10}y}\right).$

Solution:

Employ the distributive property, and then simplify the result.

$$\begin{array}{ccc}\hfill \sqrt[\mathrm{iii}]{\mathrm{six}{x}^{\mathrm{two}}y}(\sqrt[3]{9x^2{y}^2}-\mathrm{five}\cdot \sqrt[\mathrm{iii}]{4xy})& =\hfill & \sqrt[3]{6x^{2}y}\cdot \sqrt[\mathrm{three}]{9x^2{y}^{\mathrm{ii}}}-\sqrt[\mathrm{three}]{6x^{\mathrm{two}}y}\cdot 5\sqrt[\mathrm{three}]{\mathrm{four}xy}\hfill \\ \hfill & =\hfill & \sqrt[3]{54x^4{y}^{\mathrm{three}}}-5\sqrt[\mathrm{iii}]{24x^3{y}^{\mathrm{ii}}}\hfill \\ \hfill & =\hfill & \sqrt[3]{27\cdot 2\cdot \mathrm{ten}\cdot x^3\cdot y^3}-5\sqrt[3]{8\cdot \mathrm{iii}\cdot x^{\mathrm{three}}\cdot y^2}\hfill \\ \hfill & =\hfill & 3xy\sqrt[3]{2\mathrm{10}}-5\cdot 2\mathrm{10}\sqrt[3]{3y^{\mathrm{two}}}\hfill \\ \hfill & =\hfill & 3\mathrm{ten}y\sqrt[3]{2\mathrm{ten}}-\mathrm{ten}\mathrm{10}\sqrt[3]{3y^2}\hfill \end{array}$$

Answer: $\mathrm{three}\mathrm{ten}y\sqrt[3]{2\mathrm{ten}}-10\mathrm{ten}\sqrt[3]{3y^2}$

The procedure for multiplying radical expressions with multiple terms is the aforementioned procedure used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.

Example half-dozen

Multiply: ${\left(\sqrt{x}-5\sqrt{y}\right)}^2.$

Solution:

$${\left(\sqrt{x}-5\sqrt{y}\right)}^2=\left(\sqrt{\mathrm{ten}}-5\sqrt{y}\right)\left(\sqrt{x}-5\sqrt{y}\right)$$

Begin by applying the distributive property.

$$\begin{array}{cc}=\hfill & \sqrt{x}\cdot \sqrt{x}+\sqrt{\mathrm{ten}}\left(\text{−}\mathrm{v}\sqrt{y}\right)+\left(\text{−}\mathrm{v}\sqrt{y}\right)\sqrt{x}+\left(-5\sqrt{y}\right)\left(\text{−}\mathrm{five}\sqrt{y}\right)\hfill \\ =\hfill & \sqrt{x^{\mathrm{ii}}}-\mathrm{v}\sqrt{\mathrm{ten}y}-5\sqrt{xy}+25\sqrt{y^2}\hfill \\ =\hfill & x-10\sqrt{xy}+25y\hfill \end{array}$$

Answer: $\mathrm{10}-10\sqrt{\mathrm{ten}y}+25y$

The binomials $\left(a+b\right)$ and $\left(a-b\right)$ are called conjugatesThe factors $\left(a+b\right)$ and $\left(a-b\right)$ are conjugates. . When multiplying conjugate binomials the middle terms are opposites and their sum is aught.

Example 7

Multiply: $\left(\sqrt{10}+\sqrt{\mathrm{three}}\right)\left(\sqrt{\mathrm{x}}-\sqrt{3}\right).$

Solution:

Apply the distributive property, and then combine like terms.

$$\begin{array}{llll}\left(\sqrt{10}+\sqrt{\mathrm{iii}}\right)\left(\sqrt{\mathrm{x}}-\sqrt{3}\right)\hfill & =\hfill & \sqrt{10}\cdot \sqrt{\mathrm{ten}}+\sqrt{\mathrm{ten}}\left(-\sqrt{3}\right)+\sqrt{\mathrm{iii}}\cdot \sqrt{10}+\sqrt{3}\left(-\sqrt{\mathrm{iii}}\right)\hfill & Di\mathrm{south}tribute\text{.}\hfill \\ \hfill & =\hfill & \sqrt{100}-\sqrt{\mathrm{xxx}}+\sqrt{\mathrm{thirty}}-\sqrt{9}\hfill & \mathrm{Due\; south}imp\mathrm{fifty}ify\text{.}\hfill \\ \hfill & =\hfill & 10-\sqrt{\mathrm{thirty}}+\sqrt{30}-3\hfill & opposit\mathrm{eastward}saddto0\hfill \\ \hfill & =\hfill & \mathrm{ten}-3\hfill \\ \hfill & =\hfill & 7\hfill \end{array}$$

Answer: 7

It is important to note that when multiplying cohabit radical expressions, we obtain a rational expression. This is true in general

$$\begin{array}{ccc}\hfill \left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)& =& \sqrt{x^2}-\sqrt{\mathrm{ten}y}+\sqrt{xy}-\sqrt{y^{\mathrm{ii}}}\hfill \\ \hfill & =& \mathrm{10}-y\hfill \end{array}$$

Alternatively, using the formula for the difference of squares we have,

$$\begin{array}{cccc}\hfill \left(a+b\right)\left(a-b\right)& =& a^2-b^2\hfill & Diff\mathrm{east}renceofsquare\mathrm{due\; south}.\\ \hfill \left(\sqrt{\mathrm{10}}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)& =& {\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2\hfill & \\ \hfill & =& x-y\hfill & \end{array}$$

Attempt this! Multiply: $\left(3-\mathrm{two}\sqrt{y}\right)\left(3+2\sqrt{y}\right).$ (Assume y is positive.)

Answer: $\mathrm{nine}-4y$

Dividing Radical Expressions

To split up radical expressions with the aforementioned index, we use the quotient rule for radicals. Given real numbers $\sqrt[n]{A}$ and $\sqrt[\mathrm{due\; north}]{B}$ ,

$$\frac{\sqrt[n]{A}}{\sqrt[n]{B}}=\sqrt[n]{\frac{A}{B}}$$

Case 8

Divide: $\frac{\sqrt[3]{96}}{\sqrt[3]{6}}$ .

Solution:

In this case, we tin see that 6 and 96 have mutual factors. If we use the caliber rule for radicals and write it as a unmarried cube root, we will exist able to reduce the fractional radicand.

$$\begin{array}{lllll}\hfill \frac{\sqrt[3]{96}}{\sqrt[3]{\mathrm{vi}}}& =\hfill & \sqrt[\mathrm{three}]{\frac{96}{\mathrm{six}}}\hfill & \hfill & Applyth\mathrm{east}quotientrul\mathrm{east}forradica\mathrm{fifty}sa\mathrm{due\; north}dr\mathrm{east}ducetheradicand.\hfill \\ \hfill & =\hfill & \sqrt[3]{\mathrm{xvi}}\hfill & \hfill & \mathrm{South}implify.\hfill \\ \hfill & =\hfill & \sqrt[3]{8\cdot \mathrm{two}}\hfill & \hfill & \hfill \\ \hfill & =\hfill & 2\sqrt[3]{\mathrm{two}}\hfill & \hfill & \hfill \end{array}$$

Answer: $2\sqrt[\mathrm{iii}]{\mathrm{ii}}$

Example 9

Divide: $\frac{\sqrt{50x^{\mathrm{vi}}{y}^4}}{\sqrt{\mathrm{eight}{x}^{\mathrm{three}}y}}$ .

Solution:

Write as a single square root and cancel common factors before simplifying.

$$\begin{array}{lllll}\hfill \frac{\sqrt{50x^6{y}^4}}{\sqrt{8{\mathrm{ten}}^{3}y}}& =\hfill & \sqrt{\frac{\mathrm{l}{x}^6{y}^{\mathrm{iv}}}{8x^{3}y}}\hfill & \hfill & Applythequotientru\mathrm{fifty}eforradical\mathrm{southward}a\mathrm{northward}dca\mathrm{due\; north}c\mathrm{east}l.\hfill \\ \hfill & =\hfill & \sqrt{\frac{25x^3{y}^3}{4}}\hfill & \hfill & \mathrm{Due\; south}imp\mathrm{50}ify.\hfill \\ \hfill & =\hfill & \frac{\sqrt{25x^3{y}^{\mathrm{iii}}}}{\sqrt{\mathrm{four}}}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{5\mathrm{10}y\sqrt{xy}}{2}\hfill & \hfill & \hfill \end{array}$$

Answer: $\frac{5xy\sqrt{\mathrm{ten}y}}{2}$

Rationalizing the Denominator

When the denominator (divisor) of a radical expression contains a radical, it is a common practice to notice an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominatorThe process of determining an equivalent radical expression with a rational denominator. .

$$\begin{array}{ccccc}\hfill Radicale\mathrm{10}pr\mathrm{east}\mathrm{due\; south}sio\mathrm{northward}\hfill & \hfill \hfill & \hfill \hfill & \hfill Ratio\mathrm{due\; north}a\mathrm{fifty}d\mathrm{east}no\mathrm{grand}inator\hfill & \hfill \hfill \\ \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \hfill & \hfill =\hfill & \hfill \frac{\sqrt{2}}{\mathrm{two}}\hfill & \hfill \hfill \end{array}$$

To do this, multiply the fraction by a special course of 1 then that the radicand in the denominator can exist written with a power that matches the alphabetize. Later doing this, simplify and eliminate the radical in the denominator. For example:

$$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{\mathrm{ii}}$$

Recollect, to obtain an equivalent expression, y'all must multiply the numerator and denominator past the verbal same nonzero factor.

Instance 10

Rationalize the denominator: $\frac{\sqrt{2}}{\sqrt{\mathrm{five}x}}.$

Solution:

The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize information technology. In this example, multiply by 1 in the class $\frac{\sqrt{5\mathrm{ten}}}{\sqrt{5x}}.$

$$\begin{array}{lllll}\hfill \frac{\sqrt{2}}{\sqrt{5x}}& =\hfill & \frac{\sqrt{2}}{\sqrt{\mathrm{five}\mathrm{ten}}}\cdot \frac{\sqrt{5x}}{\sqrt{5\mathrm{ten}}}\hfill & \hfill & Multiplyby\frac{\sqrt{\mathrm{five}x}}{\sqrt{5x}}.\hfill \\ \hfill & =\hfill & \frac{\sqrt{10x}}{\sqrt{25x^{\mathrm{two}}}}\hfill & \hfill & Si\mathrm{thou}plify.\hfill \\ \hfill & =\hfill & \frac{\sqrt{10\mathrm{10}}}{\mathrm{v}\mathrm{10}}\hfill & \hfill & \hfill \end{array}$$

Answer: $\frac{\sqrt{10x}}{5x}$

Sometimes, we will notice the need to reduce, or abolish, subsequently rationalizing the denominator.

Example 11

Rationalize the denominator: $\frac{3a\sqrt{2}}{\sqrt{6ab}}.$

Solution:

In this case, we will multiply past 1 in the class $\frac{\sqrt{6ab}}{\sqrt{6ab}}.$

$$\begin{array}{lllll}\hfill \frac{3a\sqrt{2}}{\sqrt{6ab}}& =\hfill & \frac{3a\sqrt{2}}{\sqrt{6ab}}\cdot \frac{\sqrt{6ab}}{\sqrt{\mathrm{six}ab}}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{3a\sqrt{12ab}}{\sqrt{36a^2{b}^{\mathrm{two}}}}\hfill & \hfill & \mathrm{Southward}implify.\hfill \\ \hfill & =\hfill & \frac{3a\sqrt{\mathrm{four}\cdot 3ab}}{\mathrm{six}ab}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{6a\sqrt{\mathrm{iii}ab}}{6ab}\hfill & \hfill & Canc\mathrm{due\; east}l.\hfill \\ \hfill & =\hfill & \frac{\sqrt{3ab}}{b}\hfill & \hfill & \hfill \end{array}$$

Find that b does not cancel in this instance. Practise not cancel factors inside a radical with those that are outside.

Answer: $\frac{\sqrt{3ab}}{b}$

Endeavor this! Rationalize the denominator: $\sqrt{\frac{9x}{\mathrm{ii}y}}.$

Answer: $\frac{3\sqrt{2\mathrm{ten}y}}{2y}$

Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is truthful simply when the denominator contains a square root. However, this is not the case for a cube root. For example, $$\frac{1}{\sqrt[3]{x}}\cdot \frac{\sqrt[\mathrm{iii}]{x}}{\sqrt[3]{x}}=\frac{\sqrt[3]{x}}{\sqrt[3]{x^2}}$$ Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by 1 in the form of $\frac{\sqrt[\mathrm{three}]{x^2}}{\sqrt[\mathrm{three}]{{\mathrm{ten}}^{\mathrm{two}}}}$ , then nosotros can write the radicand in the denominator as a ability of 3. Simplifying the event then yields a rationalized denominator.

$$\frac{\mathrm{ane}}{\sqrt[3]{\mathrm{ten}}}=\frac{1}{\sqrt[\mathrm{three}]{\mathrm{10}}}\cdot \frac{\sqrt[3]{{\mathrm{ten}}^2}}{\sqrt[\mathrm{three}]{x^2}}=\frac{\sqrt[\mathrm{three}]{{\mathrm{ten}}^2}}{\sqrt[3]{x^3}}=\frac{\sqrt[3]{x^2}}{x}$$

Therefore, to rationalize the denominator of a radical expression with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the alphabetize make up one's mind what we should multiply past. Multiply the numerator and denominator by the nth root of factors that produce nthursday powers of all the factors in the radicand of the denominator.

Example 12

Rationalize the denominator: $\frac{\sqrt[3]{2}}{\sqrt[3]{25}}.$

Solution:

The radical in the denominator is equivalent to $\sqrt[3]{5^2}.$ To rationalize the denominator, we demand: $\sqrt[3]{{\mathrm{five}}^3}.$ To obtain this, nosotros need one more factor of 5. Therefore, multiply by 1 in the form of $\frac{\sqrt[\mathrm{iii}]{5}}{\sqrt[3]{\mathrm{five}}}.$

$$\begin{array}{lllll}\hfill \frac{\sqrt[3]{\mathrm{two}}}{\sqrt[\mathrm{iii}]{25}}& =\hfill & \frac{\sqrt[3]{\mathrm{two}}}{\sqrt[3]{5^{\mathrm{two}}}}\cdot \frac{\sqrt[3]{\mathrm{five}}}{\sqrt[3]{\mathrm{five}}}\hfill & \hfill & Multip\mathrm{50}ybyth\mathrm{east}cuberootoffactorsthatr\mathrm{east}\mathrm{due\; south}ultinpower\mathrm{southward}of\mathrm{three.}\hfill \\ \hfill & =\hfill & \frac{\sqrt[\mathrm{three}]{\mathrm{ten}}}{\sqrt[\mathrm{three}]{5^3}}\hfill & \hfill & Simp\mathrm{50}ify.\hfill \\ \hfill & =\hfill & \frac{\sqrt[3]{\mathrm{x}}}{5}\hfill & \hfill & \hfill \end{array}$$

Answer: $\frac{\sqrt[3]{\mathrm{x}}}{5}$

Instance xiii

Rationalize the denominator: $\sqrt[3]{\frac{27a}{\mathrm{two}{b}^2}}$ .

Solution:

In this case, we will multiply by one in the course $\frac{\sqrt[3]{2^{2}b}}{\sqrt[\mathrm{three}]{{\mathrm{two}}^{2}b}}$ .

$$\begin{array}{lllll}\hfill \sqrt[3]{\frac{27a}{2b^{\mathrm{ii}}}}& =\hfill & \frac{\sqrt[3]{3^{3}a}}{\sqrt[3]{2b^{\mathrm{two}}}}\hfill & \hfill & Applyth\mathrm{east}quotientru\mathrm{fifty}eforradicals.\hfill \\ \hfill & =\hfill & \frac{3\sqrt[3]{a}}{\sqrt[3]{2b^2}}\cdot \frac{\sqrt[3]{2^{2}b}}{\sqrt[3]{{\mathrm{ii}}^{2}b}}\hfill & \hfill & Mu\mathrm{50}tip\mathrm{50}ybyth\mathrm{eastward}cuberootoffactorsthatr\mathrm{east}sulti\mathrm{north}power\mathrm{due\; south}of\mathrm{three.}\hfill \\ \hfill & =\hfill & \frac{3\sqrt[\mathrm{iii}]{2^{2}ab}}{\sqrt[3]{2^3{b}^{\mathrm{three}}}}\hfill & \hfill & \mathrm{South}implify.\hfill \\ \hfill & =\hfill & \frac{3\sqrt[\mathrm{three}]{4ab}}{2b}\hfill & \hfill & \hfill \end{array}$$

Respond: $\frac{\mathrm{iii}\sqrt[\mathrm{iii}]{4ab}}{\mathrm{two}b}$

Example fourteen

Rationalize the denominator: $\frac{\mathrm{two}x\sqrt[5]{\mathrm{five}}}{\sqrt[5]{\mathrm{iv}{\mathrm{10}}^{3}y}}$ .

Solution:

In this example, we will multiply by 1 in the form $\frac{\sqrt[\mathrm{five}]{2^{\mathrm{iii}}{\mathrm{10}}^2{y}^{\mathrm{iv}}}}{\sqrt[5]{{\mathrm{ii}}^{\mathrm{iii}}{\mathrm{10}}^2{y}^{\mathrm{iv}}}}$ .

$$\begin{array}{lllll}\hfill \frac{2\mathrm{ten}\sqrt[5]{5}}{\sqrt[\mathrm{v}]{4x^{\mathrm{three}}y}}& =\hfill & \frac{2x\sqrt[5]{5}}{\sqrt[5]{2^2{x}^{\mathrm{iii}}y}}\cdot \frac{\sqrt[\mathrm{five}]{{\mathrm{ii}}^3{x}^2{y}^4}}{\sqrt[5]{2^{\mathrm{iii}}{\mathrm{ten}}^{\mathrm{two}}{y}^{\mathrm{four}}}}\hfill & \hfill & \mathrm{Thousand}u\mathrm{fifty}tiplybyth\mathrm{eastward}fifthrootoffactor\mathrm{southward}thatr\mathrm{due\; east}su\mathrm{fifty}ti\mathrm{north}power\mathrm{southward}of5.\hfill \\ \hfill & =\hfill & \frac{2\mathrm{10}\sqrt[\mathrm{five}]{5\cdot 2^3{x}^2{y}^4}}{\sqrt[\mathrm{v}]{{\mathrm{ii}}^5{x}^5{y}^5}}\hfill & \hfill & Simplify.\hfill \\ \hfill & =\hfill & \frac{2\mathrm{ten}\sqrt[5]{\mathrm{twoscore}{x}^2{y}^4}}{2xy}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\sqrt[5]{\mathrm{twoscore}{\mathrm{ten}}^{\mathrm{ii}}{y}^{\mathrm{four}}}}{y}\hfill & \hfill & \hfill \end{array}$$

Answer: $\frac{\sqrt[\mathrm{v}]{\mathrm{twoscore}{x}^2{y}^{\mathrm{iv}}}}{y}$

When two terms involving square roots appear in the denominator, we can rationalize information technology using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction past the conjugate of the denominator. Recall that multiplying a radical expression by its cohabit produces a rational number.

Example 15

Rationalize the denominator: $\frac{\mathrm{one}}{\sqrt{\mathrm{five}}-\sqrt{3}}.$

Solution:

In this example, the cohabit of the denominator is $\sqrt{\mathrm{five}}+\sqrt{3}.$ Therefore, multiply by 1 in the grade $\frac{\left(\sqrt{5}+\sqrt{\mathrm{three}}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}.$

$$\begin{array}{cccc}\hfill \frac{1}{\sqrt{5}-\sqrt{3}}& =\hfill & \frac{1}{\left(\sqrt{\mathrm{v}}-\sqrt{3}\right)}\frac{\left(\sqrt{\mathrm{v}}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}\hfill & \begin{array}{c}\\ \\ \\ \\ \mathrm{1000}ultiply\mathrm{northward}um\mathrm{due\; east}ratora\mathrm{due\; north}d\hfill \\ d\mathrm{eastward}nogi\mathrm{northward}atorbyth\mathrm{east}conjukate\hfill \\ ofthede\mathrm{due\; north}o\mathrm{yard}i\mathrm{due\; north}ator.\hfill \end{array}\hfill \\ \hfill & =\hfill & \frac{\sqrt{\mathrm{five}}+\sqrt{3}}{\sqrt{25}+\sqrt{\mathrm{fifteen}}-\sqrt{15}-\sqrt{9}}\hfill & Simplify\text{.}\hfill \\ \hfill & =\hfill & \frac{\sqrt{\mathrm{five}}+\sqrt{3}}{5-3}\hfill & \hfill \\ \hfill & =\hfill & \frac{\sqrt{5}+\sqrt{3}}{\mathrm{two}}\hfill & \hfill \end{array}$$

Reply: $\frac{\sqrt{\mathrm{five}}+\sqrt{\mathrm{three}}}{2}$

Observe that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can apply the property $\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b$ to expedite the process of multiplying the expressions in the denominator.

Example sixteen

Rationalize the denominator: $\frac{\sqrt{\mathrm{ten}}}{\sqrt{\mathrm{ii}}+\sqrt{\mathrm{vi}}}$ .

Solution:

Multiply past 1 in the form $\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}$ .

$$\begin{array}{lllll}\hfill \frac{\sqrt{\mathrm{x}}}{\sqrt{2}+\sqrt{\mathrm{half\; dozen}}}& =\hfill & \frac{\left(\sqrt{10}\right)}{\left(\sqrt{2}+\sqrt{\mathrm{half-dozen}}\right)}\frac{\left(\sqrt{2}-\sqrt{6}\right)}{\left(\sqrt{2}-\sqrt{6}\right)}\hfill & \hfill & Multiplybyth\mathrm{eastward}conju\mathrm{thou}at\mathrm{eastward}ofthede\mathrm{northward}oginator.\hfill \\ \hfill & =\hfill & \frac{\sqrt{\mathrm{xx}}-\sqrt{\mathrm{lx}}}{2-\mathrm{half-dozen}}\hfill & \hfill & Simplify.\hfill \\ \hfill & =\hfill & \frac{\sqrt{4\cdot \mathrm{five}}-\sqrt{\mathrm{four}\cdot 15}}{\text{−}4}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{2\sqrt{5}-2\sqrt{\mathrm{fifteen}}}{\text{−}4}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\mathrm{two}\left(\sqrt{5}-\sqrt{15}\right)}{\text{−}4}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\sqrt{5}-\sqrt{15}}{\text{−}2}=-\frac{\sqrt{5}-\sqrt{15}}{2}=\frac{-\sqrt{\mathrm{five}}+\sqrt{15}}{\mathrm{two}}\hfill & \hfill & \hfill \end{array}$$

Answer: $\frac{\sqrt{15}-\sqrt{5}}{\mathrm{two}}$

Example 17

Rationalize the denominator: $\frac{\sqrt{\mathrm{ten}}-\sqrt{y}}{\sqrt{\mathrm{ten}}+\sqrt{y}}$ .

Solution:

In this example, we volition multiply by 1 in the course $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}.$

$$\begin{array}{lllll}\hfill \frac{\sqrt{x}-\sqrt{y}}{\sqrt{\mathrm{10}}+\sqrt{y}}& =\hfill & \frac{\left(\sqrt{\mathrm{ten}}-\sqrt{y}\right)}{\left(\sqrt{\mathrm{10}}+\sqrt{y}\right)}\frac{\left(\sqrt{\mathrm{10}}-\sqrt{y}\right)}{\left(\sqrt{\mathrm{ten}}-\sqrt{y}\right)}\hfill & \hfill & Multiplybytheco\mathrm{due\; north}jumateofthed\mathrm{due\; east}\mathrm{northward}oki\mathrm{northward}ator.\hfill \\ \hfill & =\hfill & \frac{\sqrt{x^{\mathrm{ii}}}-\sqrt{xy}-\sqrt{xy}+\sqrt{y^2}}{\mathrm{10}-y}\hfill & \hfill & Si\mathrm{1000}p\mathrm{fifty}ify.\hfill \\ \hfill & =\hfill & \frac{x-2\sqrt{xy}+y}{x-y}\hfill & \hfill & \hfill \end{array}$$

Reply: $\frac{x-2\sqrt{\mathrm{10}y}+y}{\mathrm{10}-y}$

Try this! Rationalize the denominator: $\frac{\mathrm{ii}\sqrt{3}}{5-\sqrt{\mathrm{iii}}}$

Answer: $\frac{5\sqrt{\mathrm{three}}+3}{11}$

Key Takeaways

• To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result.
• Use the distributive holding when multiplying a radical expression with multiple terms. Then simplify and combine all like radicals.
• Multiplying a ii-term radical expression involving square roots by its conjugate results in a rational expression.
• Information technology is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator.
• If an expression has one term in the denominator involving a radical, then rationalize it by multiplying the numerator and denominator by the nth root of factors of the radicand so that their powers equal the index.
• If a radical expression has ii terms in the denominator involving square roots, so rationalize it by multiplying the numerator and denominator by the conjugate of the denominator.

Topic Exercises

Part A: Multiplying Radical Expressions

Multiply. (Assume all variables represent non-negative real numbers.)

1. $\sqrt{3}\cdot \sqrt{7}$

2. $\sqrt{2}\cdot \sqrt{5}$

3. $\sqrt{6}\cdot \sqrt{12}$

4. $\sqrt{10}\cdot \sqrt{15}$

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

6. $\sqrt{5}\cdot \sqrt{\mathrm{fifteen}}$

7. $\sqrt{\mathrm{vii}}\cdot \sqrt{7}$

8. $\sqrt{12}\cdot \sqrt{12}$

9. $2\sqrt{5}\cdot 7\sqrt{10}$

10. $3\sqrt{15}\cdot 2\sqrt{\mathrm{half\; dozen}}$

11. ${\left(2\sqrt{\mathrm{five}}\right)}^2$

12. ${\left(6\sqrt{\mathrm{two}}\right)}^{\mathrm{two}}$

13. $\sqrt{2\mathrm{ten}}\cdot \sqrt{2x}$

14. $\sqrt{5y}\cdot \sqrt{5y}$

15. $\sqrt{3a}\cdot \sqrt{12}$

16. $\sqrt{\mathrm{three}a}\cdot \sqrt{\mathrm{ii}a}$

17. $4\sqrt{\mathrm{two}\mathrm{ten}}\cdot 3\sqrt{\mathrm{half\; dozen}\mathrm{ten}}$

18. $\mathrm{v}\sqrt{10y}\cdot \mathrm{two}\sqrt{2y}$

19. $\sqrt[3]{\mathrm{iii}}\cdot \sqrt[\mathrm{three}]{9}$

20. $\sqrt[\mathrm{three}]{\mathrm{iv}}\cdot \sqrt[\mathrm{iii}]{\mathrm{xvi}}$

21. $\sqrt[\mathrm{three}]{\mathrm{fifteen}}\cdot \sqrt[3]{25}$

22. $\sqrt[\mathrm{three}]{100}\cdot \sqrt[3]{50}$

23. $\sqrt[3]{\mathrm{iv}}\cdot \sqrt[3]{10}$

24. $\sqrt[\mathrm{three}]{18}\cdot \sqrt[3]{6}$

25. $\left(5\sqrt[\mathrm{iii}]{9}\right)\left(2\sqrt[3]{6}\right)$

26. $\left(\mathrm{ii}\sqrt[3]{\mathrm{iv}}\right)\left(\mathrm{three}\sqrt[3]{4}\right)$

27. ${\left(2\sqrt[3]{2}\right)}^3$

28. ${\left(3\sqrt[\mathrm{iii}]{4}\right)}^{\mathrm{iii}}$

29. $\sqrt[3]{3a^2}\cdot \sqrt[3]{9a}$

30. $\sqrt[3]{7b}\cdot \sqrt[3]{49b^{\mathrm{ii}}}$

31. $\sqrt[3]{\mathrm{half\; dozen}{x}^2}\cdot \sqrt[\mathrm{three}]{4x^{\mathrm{two}}}$

32. $\sqrt[\mathrm{iii}]{12y}\cdot \sqrt[3]{9y^2}$

33. $\sqrt[3]{20{\mathrm{10}}^{2}y}\cdot \sqrt[\mathrm{three}]{\mathrm{x}{\mathrm{ten}}^{\mathrm{two}}{y}^2}$

34. $\sqrt[3]{63\mathrm{10}y}\cdot \sqrt[3]{12{\mathrm{ten}}^4{y}^2}$

35. $\sqrt{\mathrm{v}}\left(3-\sqrt{5}\right)$

36. $\sqrt{2}\left(\sqrt{\mathrm{iii}}-\sqrt{\mathrm{two}}\right)$

37. $3\sqrt{7}\left(\mathrm{ii}\sqrt{7}-\sqrt{3}\right)$

38. $2\sqrt{\mathrm{v}}\left(6-\mathrm{iii}\sqrt{10}\right)$

39. $\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)$

40. $\sqrt{15}\left(\sqrt{\mathrm{v}}+\sqrt{\mathrm{iii}}\right)$

41. $\sqrt{x}\left(\sqrt{\mathrm{10}}+\sqrt{xy}\right)$

42. $\sqrt{y}\left(\sqrt{\mathrm{ten}y}+\sqrt{y}\right)$

43. $\sqrt{\mathrm{ii}ab}\left(\sqrt{14a}-2\sqrt{10b}\right)$

44. $\sqrt{\mathrm{six}ab}\left(\mathrm{five}\sqrt{2a}-\sqrt{3b}\right)$

45. $\sqrt[3]{6}\left(\sqrt[3]{\mathrm{nine}}-\sqrt[\mathrm{iii}]{20}\right)$

46. $\sqrt[3]{12}\left(\sqrt[\mathrm{iii}]{36}+\sqrt[3]{14}\right)$

47. $\left(\sqrt{\mathrm{two}}-\sqrt{\mathrm{v}}\right)\left(\sqrt{\mathrm{iii}}+\sqrt{7}\right)$

48. $\left(\sqrt{3}+\sqrt{\mathrm{ii}}\right)\left(\sqrt{5}-\sqrt{7}\right)$

49. $\left(2\sqrt{3}-\mathrm{iv}\right)\left(3\sqrt{6}+1\right)$

50. $\left(\mathrm{v}-2\sqrt{\mathrm{six}}\right)\left(7-2\sqrt{\mathrm{three}}\right)$

51. ${\left(\sqrt{\mathrm{v}}-\sqrt{\mathrm{iii}}\right)}^{\mathrm{ii}}$

52. ${\left(\sqrt{\mathrm{vii}}-\sqrt{2}\right)}^2$

53. $\left(2\sqrt{\mathrm{iii}}+\sqrt{2}\right)\left(2\sqrt{\mathrm{iii}}-\sqrt{\mathrm{two}}\right)$

54. $\left(\sqrt{\mathrm{ii}}+3\sqrt{7}\right)\left(\sqrt{\mathrm{two}}-\mathrm{iii}\sqrt{7}\right)$

55. ${\left(\sqrt{a}-\sqrt{2b}\right)}^2$

56. ${\left(\sqrt{ab}+1\right)}^2$

57. What is the perimeter and area of a rectangle with length measuring $5\sqrt{3}$ centimeters and width measuring $\mathrm{three}\sqrt{2}$ centimeters?

58. What is the perimeter and surface area of a rectangle with length measuring $2\sqrt{\mathrm{half-dozen}}$ centimeters and width measuring $\sqrt{3}$ centimeters?

59. If the base of operations of a triangle measures $\mathrm{half-dozen}\sqrt{2}$ meters and the height measures $3\sqrt{\mathrm{two}}$ meters, so calculate the area.

60. If the base of a triangle measures $6\sqrt{3}$ meters and the height measures $3\sqrt{6}$ meters, then calculate the surface area.

Part B: Dividing Radical Expressions

Divide. (Assume all variables represent positive real numbers.)

61. $$\frac{\sqrt{75}}{\sqrt{\mathrm{three}}}$$

62. $$\frac{\sqrt{360}}{\sqrt{\mathrm{ten}}}$$

63. $$\frac{\sqrt{72}}{\sqrt{75}}$$

64. $$\frac{\sqrt{90}}{\sqrt{98}}$$

65. $$\frac{\sqrt{90{\mathrm{ten}}^5}}{\sqrt{2\mathrm{10}}}$$

66. $$\frac{\sqrt{96y^3}}{\sqrt{3y}}$$

67. $$\frac{\sqrt{162{\mathrm{10}}^7{y}^{\mathrm{v}}}}{\sqrt{\mathrm{two}\mathrm{ten}y}}$$

68. $$\frac{\sqrt{363x^4{y}^9}}{\sqrt{3xy}}$$

69. $$\frac{\sqrt[3]{16a^5{b}^2}}{\sqrt[3]{2a^2{b}^2}}$$

70. $$\frac{\sqrt[3]{192a^{\mathrm{two}}{b}^7}}{\sqrt[3]{2a^2{b}^2}}$$

Part C: Rationalizing the Denominator

Rationalize the denominator. (Assume all variables represent positive real numbers.)

71. $$\frac{\mathrm{ane}}{\sqrt{5}}$$

72. $$\frac{1}{\sqrt{\mathrm{vi}}}$$

73. $$\frac{\sqrt{2}}{\sqrt{3}}$$

74. $$\frac{\sqrt{\mathrm{iii}}}{\sqrt{\mathrm{vii}}}$$

75. $$\frac{\mathrm{five}}{2\sqrt{10}}$$

76. $$\frac{\mathrm{three}}{5\sqrt{6}}$$

77. $$\frac{\sqrt{\mathrm{three}}-\sqrt{5}}{\sqrt{\mathrm{three}}}$$

78. $$\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}}$$

79. $$\frac{1}{\sqrt{7\mathrm{ten}}}$$

80. $$\frac{\mathrm{i}}{\sqrt{3y}}$$

81. $$\frac{a}{5\sqrt{ab}}$$

82. $$\frac{\mathrm{three}{b}^2}{2\sqrt{\mathrm{three}ab}}$$

83. $$\frac{\mathrm{ii}}{\sqrt[\mathrm{three}]{36}}$$

84. $$\frac{\mathrm{xiv}}{\sqrt[3]{7}}$$

85. $$\frac{1}{\sqrt[3]{\mathrm{four}\mathrm{ten}}}$$

86. $$\frac{1}{\sqrt[3]{3y^2}}$$

87. $$\frac{9x\sqrt[3]{\mathrm{ii}}}{\sqrt[3]{9x{y}^{\mathrm{two}}}}$$

88. $$\frac{5y^2\sqrt[3]{x}}{\sqrt[3]{5x^{2}y}}$$

89. $$\frac{\mathrm{iii}a}{2\sqrt[\mathrm{iii}]{\mathrm{three}{a}^{\mathrm{ii}}{b}^2}}$$

90. $$\frac{25n}{\mathrm{three}\sqrt[\mathrm{three}]{25{\mathrm{chiliad}}^{\mathrm{two}}\mathrm{due\; north}}}$$

91. $$\frac{3}{\sqrt[\mathrm{v}]{27x^{2}y}}$$

92. $$\frac{2}{\sqrt[\mathrm{five}]{16x{y}^2}}$$

93. $$\frac{ab}{\sqrt[\mathrm{five}]{9a^{3}b}}$$

94. $$\frac{abc}{\sqrt[\mathrm{v}]{a{b}^2{c}^{\mathrm{three}}}}$$

95. $$\sqrt[\mathrm{five}]{\frac{\mathrm{three}x}{\mathrm{viii}{y}^{2}z}}$$

96. $$\sqrt[5]{\frac{\mathrm{four}x{y}^2}{\mathrm{ix}{x}^{\mathrm{three}}y{z}^{\mathrm{four}}}}$$

97. $$\frac{\mathrm{iii}}{\sqrt{10}-3}$$

98. $$\frac{2}{\sqrt{6}-\mathrm{ii}}$$

99. $$\frac{\mathrm{ane}}{\sqrt{5}+\sqrt{3}}$$

100. $$\frac{1}{\sqrt{7}-\sqrt{2}}$$

101. $$\frac{\sqrt{3}}{\sqrt{\mathrm{three}}+\sqrt{\mathrm{half-dozen}}}$$

102. $$\frac{\sqrt{5}}{\sqrt{5}+\sqrt{15}}$$

103. $$\frac{\mathrm{x}}{5-3\sqrt{5}}$$

104. $$\frac{\text{−}2\sqrt{\mathrm{two}}}{4-3\sqrt{2}}$$

105. $$\frac{\sqrt{\mathrm{iii}}+\sqrt{5}}{\sqrt{3}-\sqrt{5}}$$

106. $$\frac{\sqrt{\mathrm{x}}-\sqrt{2}}{\sqrt{\mathrm{x}}+\sqrt{2}}$$

107. $$\frac{2\sqrt{3}-3\sqrt{\mathrm{two}}}{4\sqrt{\mathrm{three}}+\sqrt{2}}$$

108. $$\frac{6\sqrt{5}+\mathrm{two}}{\mathrm{ii}\sqrt{\mathrm{five}}-\sqrt{\mathrm{ii}}}$$

109. $$\frac{x-y}{\sqrt{x}+\sqrt{y}}$$

110. $$\frac{x-y}{\sqrt{\mathrm{ten}}-\sqrt{y}}$$

111. $$\frac{\mathrm{10}+\sqrt{y}}{x-\sqrt{y}}$$

112. $$\frac{x-\sqrt{y}}{x+\sqrt{y}}$$

113. $$\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$$

114. $$\frac{\sqrt{ab}+\sqrt{2}}{\sqrt{ab}-\sqrt{2}}$$

115. $$\frac{\sqrt{x}}{5-2\sqrt{\mathrm{10}}}$$

116. $$\frac{\mathrm{i}}{\sqrt{x}-y}$$

117. $$\frac{\sqrt{\mathrm{10}}+\sqrt{2y}}{\sqrt{2x}-\sqrt{y}}$$

118. $$\frac{\sqrt{3x}-\sqrt{y}}{\sqrt{x}+\sqrt{3y}}$$

119. $$\frac{\sqrt{2x+1}}{\sqrt{2x+\mathrm{ane}}-1}$$

120. $$\frac{\sqrt{x+1}}{1-\sqrt{\mathrm{ten}+1}}$$

121. $$\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{\mathrm{10}-1}}$$

122. $$\frac{\sqrt{\mathrm{two}x+\mathrm{iii}}-\sqrt{2x-\mathrm{three}}}{\sqrt{\mathrm{two}\mathrm{ten}+3}+\sqrt{\mathrm{two}x-3}}$$

123. The radius of the base of a right circular cone is given past $r=\sqrt{\frac{\mathrm{three}V}{\pi h}}$ where V represents the volume of the cone and h represents its summit. Find the radius of a right circular cone with book 50 cubic centimeters and acme 4 centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.

124. The radius of a sphere is given by $r=\sqrt[3]{\frac{3V}{4\pi }}$ where V represents the volume of the sphere. Find the radius of a sphere with volume 135 square centimeters. Give the exact answer and the judge reply rounded to the nearest hundredth.

Part D: Discussion

125. Research and discuss some of the reasons why information technology is a common practise to rationalize the denominator.

126. Explain in your ain words how to rationalize the denominator.

Answers

1. $\sqrt{21}$

3. $6\sqrt{2}$

5. $2\sqrt{\mathrm{iii}}$

7. 7

9. $70\sqrt{2}$

11. twenty

13. $2x$

15. $6\sqrt{a}$

17. $24\mathrm{ten}\sqrt{\mathrm{three}}$

19. three

21. $5\sqrt[3]{3}$

23. $\mathrm{ii}\sqrt[3]{5}$

25. $\mathrm{xxx}\sqrt[3]{\mathrm{two}}$

27. 16

29. $\mathrm{iii}a$

31. $2\mathrm{ten}\sqrt[3]{3x}$

33. $2xy\sqrt[3]{25x}$

35. $3\sqrt{5}-\mathrm{v}$

37. $42-3\sqrt{21}$

39. $\mathrm{three}\sqrt{2}-\mathrm{two}\sqrt{3}$

41. $\mathrm{ten}+x\sqrt{y}$

43. $\mathrm{two}a\sqrt{7b}-4b\sqrt{\mathrm{v}a}$

45. $3\sqrt[3]{2}-2\sqrt[\mathrm{three}]{15}$

47. $\sqrt{6}+\sqrt{14}-\sqrt{15}-\sqrt{35}$

49. $18\sqrt{\mathrm{ii}}+2\sqrt{3}-12\sqrt{\mathrm{half\; dozen}}-4$

51. $\mathrm{eight}-\mathrm{ii}\sqrt{15}$

53. 10

55. $a-\mathrm{ii}\sqrt{\mathrm{ii}ab}+2b$

57. Perimeter: $\left(10\sqrt{3}+6\sqrt{\mathrm{ii}}\right)$ centimeters; area: $\mathrm{fifteen}\sqrt{6}$ square centimeters

59. $\mathrm{eighteen}$ square meters

61. five

63. $$\frac{2\sqrt{\mathrm{half\; dozen}}}{5}$$

65. $\mathrm{three}{x}^2\sqrt{5}$

67. $9{\mathrm{10}}^3{y}^2$

69. $2a$

71. $$\frac{\sqrt{5}}{5}$$

73. $$\frac{\sqrt{6}}{\mathrm{iii}}$$

75. $$\frac{\sqrt{10}}{\mathrm{four}}$$

77. $$\frac{3-\sqrt{15}}{\mathrm{three}}$$

79. $$\frac{\sqrt{7x}}{7x}$$

81. $$\frac{\sqrt{ab}}{\mathrm{v}b}$$

83. $$\frac{\sqrt[\mathrm{iii}]{6}}{3}$$

85. $$\frac{\sqrt[3]{2{\mathrm{10}}^{\mathrm{ii}}}}{2x}$$

87. $$\frac{3\sqrt[3]{6x^{\mathrm{two}}y}}{y}$$

89. $$\frac{\sqrt[\mathrm{three}]{9ab}}{2b}$$

91. $$\frac{\sqrt[\mathrm{v}]{\mathrm{nine}{x}^3{y}^4}}{xy}$$

93. $$\frac{\sqrt[5]{27a^2{b}^4}}{3}$$

95. $$\frac{\sqrt[\mathrm{five}]{12x{y}^3{z}^{\mathrm{iv}}}}{2yz}$$

97. $\mathrm{iii}\sqrt{10}+\mathrm{nine}$

99. $$\frac{\sqrt{5}-\sqrt{3}}{\mathrm{two}}$$

101. $\text{−}1+\sqrt{2}$

103. $$\frac{\text{−}5-3\sqrt{5}}{2}$$

105. $\text{−}4-\sqrt{15}$

107. $$\frac{15-7\sqrt{\mathrm{half-dozen}}}{23}$$

109. $\sqrt{x}-\sqrt{y}$

111. $$\frac{x^{\mathrm{ii}}+2x\sqrt{y}+y}{{\mathrm{10}}^{\mathrm{ii}}-y}$$

113. $$\frac{a-\mathrm{ii}\sqrt{ab}+b}{a-b}$$

115. $$\frac{\mathrm{v}\sqrt{x}+2x}{25-4x}$$

117. $$\frac{x\sqrt{2}+3\sqrt{xy}+y\sqrt{2}}{2x-y}$$

119. $$\frac{2\mathrm{ten}+1+\sqrt{2x+1}}{2\mathrm{10}}$$

121. $x+\sqrt{x^2-1}$

123. $\frac{5\sqrt{6\pi }}{2\pi }$ centimeters; 3.45 centimeters

125. Answer may vary

Multiplying And Dividing Radical Expressions,

Source: https://saylordotorg.github.io/text_intermediate-algebra/s08-04-multiplying-and-dividing-radic.html

Posted by: ridleyrenec1960.blogspot.com

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