How to divide square roots - SSAT Upper Level Quantitative

Card 0 of 32

Divide:

$18\textrm{ lbs }2\textrm{ oz } \div 5$

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We can write 18 pounds, 2 ounces as just ounces as follows:

$18\textrm{ lbs }2\textrm{ oz } = (18 \times 16 + 2) \textrm{ oz } = 290 \textrm{ oz }$

Divide:

$290 \div 5 = 58$

Convert 58 ounces to pounds and ounces as follows:

$58 \div 16 = 3 \textrm{ R }10$

That is, 3 pounds 10 ounces.

Simplify by rationalizing the denominator:

$$\frac{16}{4-\sqrt{6}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $4+$\sqrt{6}$$ :

$$\frac{16}{4-\sqrt{6}$}= $\frac{16(4+\sqrt{6}$)}{(4-$\sqrt{6}$) (4+$\sqrt{6}$)}$

$= $$\frac{16(4+\sqrt{6}$)}{4^{2}$$-($\sqrt{6}$)^{2}$ }$

$= $\frac{16(4+\sqrt{6}$)}{16-6}$

$= $\frac{16(4+\sqrt{6}$)}{10}$

$= $\frac{8(4+\sqrt{6}$)}{5}$

$= $\frac{ 32+8\sqrt{6}$}{5}$

Simplify by rationalizing the denominator:

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $6+$\sqrt{3}$$ :

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

$= $\frac{\sqrt{6}$(6+$\sqrt{3}$)}{(6-$\sqrt{3}$)(6+$\sqrt{3}$)}$

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

$= $\frac{6 \sqrt{6}$+ $\sqrt{18}$}{36-3 }$

$= $\frac{6 \sqrt{6}$+ $\sqrt{9}$ \cdot $\sqrt{2}$}{3 3 }$

$= $\frac{6 \sqrt{6}$+3$\sqrt{2}$}{3 3 }$

$= $\frac{2 \sqrt{6}$+$\sqrt{2}$}{11 }$

$= $\frac{\sqrt{2}$+ 2 $\sqrt{6}$}{11 }$

Simplify by rationalizing the denominator:

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

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Multiply both numerator and denominator by $\sqrt{6}$ :

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

$= $\frac{(3\sqrt{2}$+2$\sqrt{3}$)\cdot $\sqrt{6}$}{$\sqrt{6}$\cdot $\sqrt{6}$}$

$= $\frac{3\sqrt{2}$\cdot $\sqrt{6}$+2$\sqrt{3}$\cdot $\sqrt{6}$}{6}$

$= $\frac{3\sqrt{12}$ +2$\sqrt{18}$ }{6}$

$= $\frac{3\cdot \sqrt{4}$ \cdot $\sqrt{3}$+2\cdot $\sqrt{9}$\cdot $\sqrt{2}$ }{6}$

$= $\frac{3\cdot 2 \cdot \sqrt{3}$+2\cdot3 \cdot $\sqrt{2}$ }{6}$

$= $\frac{6 \cdot \sqrt{3}$+6 \cdot $\sqrt{2}$ }{6}$

$= $\sqrt{3}$+ $\sqrt{2}$$

Simplify by rationalizing the denominator:

$$\frac{12}{\sqrt{14}$-$\sqrt{5}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{14}$+$\sqrt{5}$ :

$$\frac{12}{\sqrt{14}$-$\sqrt{5}$}$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{($\sqrt{14}$-$\sqrt{5}$)($\sqrt{14}$+$\sqrt{5}$)}$

$= $$\frac{12(\sqrt{14}$+$\sqrt{5}$)}{($\sqrt{14}$)^{2}$$-($\sqrt{5}$)^{2}$ }$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{14-5}$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{9}$

$= $\frac{4(\sqrt{14}$+$\sqrt{5}$)}{3}$

$= $\frac{4 \sqrt{14}$+4$\sqrt{5}$}{3}$

Simplify by rationalizing the denominator:

$$\frac{6}{\sqrt{5}$+$\sqrt{7}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}$-$\sqrt{7}$ :

$$\frac{6}{\sqrt{5}$+$\sqrt{7}$}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{($\sqrt{5}$+$\sqrt{7}$)($\sqrt{5}$-$\sqrt{7}$)}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{5-7}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{-2}$

$= -3($\sqrt{5}$-$\sqrt{7}$ )$

$= 3$\sqrt{7}$-3$\sqrt{5}$$

This is not among the given choices.

Simplify by rationalizing the denominator:

$$\frac{\sqrt{15}$}{$\sqrt{5}$-$\sqrt{3}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}$+$\sqrt{3}$

$$\frac{\sqrt{15}$}{$\sqrt{5}$-$\sqrt{3}$}$

$=\frac{\sqrt{15}$( $\sqrt{5}$+$\sqrt{3}$)}{\left($\sqrt{5}$-$\sqrt{3}$ \right)( $\sqrt{5}$+$\sqrt{3}$)}$

$=\frac{ \sqrt{15}$ \cdot $\sqrt{5}$+ $\sqrt{15}$ \cdot $\sqrt{3}$}{ $($\sqrt{5}$)^{2}$-($\sqrt{3}$ $)^{2}$ }$

$=\frac{ \sqrt{15 \cdot 5}$+ $\sqrt{15 \cdot 3}$}{5-3 }$

$=\frac{ \sqrt{75}$+ $\sqrt{45}$}{2 }$

$=\frac{ \sqrt{25}$ \cdot \sqrt {3}+ $\sqrt{9}$ \cdot $\sqrt{5}$}{2 }$

$=\frac{ 5\sqrt {3}$+3 $\sqrt{5}$}{2 }$

Divide: $$\frac{\sqrt{64}$}{$\sqrt{8}$}$

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The values are of the same power. Divide the terms.

$$\frac{\sqrt{64}$}{$\sqrt{8}$}= $\sqrt{\frac{64}{8}$}= $\sqrt{8}$ = $\sqrt{4}$\times $\sqrt{2}$= 2\sqrt2$

The correct answer is: $2\sqrt2$

Divide:

$18\textrm{ lbs }2\textrm{ oz } \div 5$

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We can write 18 pounds, 2 ounces as just ounces as follows:

$18\textrm{ lbs }2\textrm{ oz } = (18 \times 16 + 2) \textrm{ oz } = 290 \textrm{ oz }$

Divide:

$290 \div 5 = 58$

Convert 58 ounces to pounds and ounces as follows:

$58 \div 16 = 3 \textrm{ R }10$

That is, 3 pounds 10 ounces.

Simplify by rationalizing the denominator:

$$\frac{16}{4-\sqrt{6}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $4+$\sqrt{6}$$ :

$$\frac{16}{4-\sqrt{6}$}= $\frac{16(4+\sqrt{6}$)}{(4-$\sqrt{6}$) (4+$\sqrt{6}$)}$

$= $$\frac{16(4+\sqrt{6}$)}{4^{2}$$-($\sqrt{6}$)^{2}$ }$

$= $\frac{16(4+\sqrt{6}$)}{16-6}$

$= $\frac{16(4+\sqrt{6}$)}{10}$

$= $\frac{8(4+\sqrt{6}$)}{5}$

$= $\frac{ 32+8\sqrt{6}$}{5}$

Simplify by rationalizing the denominator:

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $6+$\sqrt{3}$$ :

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

$= $\frac{\sqrt{6}$(6+$\sqrt{3}$)}{(6-$\sqrt{3}$)(6+$\sqrt{3}$)}$

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

$= $\frac{6 \sqrt{6}$+ $\sqrt{18}$}{36-3 }$

$= $\frac{6 \sqrt{6}$+ $\sqrt{9}$ \cdot $\sqrt{2}$}{3 3 }$

$= $\frac{6 \sqrt{6}$+3$\sqrt{2}$}{3 3 }$

$= $\frac{2 \sqrt{6}$+$\sqrt{2}$}{11 }$

$= $\frac{\sqrt{2}$+ 2 $\sqrt{6}$}{11 }$

Simplify by rationalizing the denominator:

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

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Multiply both numerator and denominator by $\sqrt{6}$ :

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

$= $\frac{(3\sqrt{2}$+2$\sqrt{3}$)\cdot $\sqrt{6}$}{$\sqrt{6}$\cdot $\sqrt{6}$}$

$= $\frac{3\sqrt{2}$\cdot $\sqrt{6}$+2$\sqrt{3}$\cdot $\sqrt{6}$}{6}$

$= $\frac{3\sqrt{12}$ +2$\sqrt{18}$ }{6}$

$= $\frac{3\cdot \sqrt{4}$ \cdot $\sqrt{3}$+2\cdot $\sqrt{9}$\cdot $\sqrt{2}$ }{6}$

$= $\frac{3\cdot 2 \cdot \sqrt{3}$+2\cdot3 \cdot $\sqrt{2}$ }{6}$

$= $\frac{6 \cdot \sqrt{3}$+6 \cdot $\sqrt{2}$ }{6}$

$= $\sqrt{3}$+ $\sqrt{2}$$

Simplify by rationalizing the denominator:

$$\frac{12}{\sqrt{14}$-$\sqrt{5}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{14}$+$\sqrt{5}$ :

$$\frac{12}{\sqrt{14}$-$\sqrt{5}$}$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{($\sqrt{14}$-$\sqrt{5}$)($\sqrt{14}$+$\sqrt{5}$)}$

$= $$\frac{12(\sqrt{14}$+$\sqrt{5}$)}{($\sqrt{14}$)^{2}$$-($\sqrt{5}$)^{2}$ }$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{14-5}$

$= $\frac{12(\sqrt{14}$+$\sqrt{5}$)}{9}$

$= $\frac{4(\sqrt{14}$+$\sqrt{5}$)}{3}$

$= $\frac{4 \sqrt{14}$+4$\sqrt{5}$}{3}$

Simplify by rationalizing the denominator:

$$\frac{6}{\sqrt{5}$+$\sqrt{7}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}$-$\sqrt{7}$ :

$$\frac{6}{\sqrt{5}$+$\sqrt{7}$}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{($\sqrt{5}$+$\sqrt{7}$)($\sqrt{5}$-$\sqrt{7}$)}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{5-7}$

$=\frac{6(\sqrt{5}$-$\sqrt{7}$)}{-2}$

$= -3($\sqrt{5}$-$\sqrt{7}$ )$

$= 3$\sqrt{7}$-3$\sqrt{5}$$

This is not among the given choices.

Simplify by rationalizing the denominator:

$$\frac{\sqrt{15}$}{$\sqrt{5}$-$\sqrt{3}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}$+$\sqrt{3}$

$$\frac{\sqrt{15}$}{$\sqrt{5}$-$\sqrt{3}$}$

$=\frac{\sqrt{15}$( $\sqrt{5}$+$\sqrt{3}$)}{\left($\sqrt{5}$-$\sqrt{3}$ \right)( $\sqrt{5}$+$\sqrt{3}$)}$

$=\frac{ \sqrt{15}$ \cdot $\sqrt{5}$+ $\sqrt{15}$ \cdot $\sqrt{3}$}{ $($\sqrt{5}$)^{2}$-($\sqrt{3}$ $)^{2}$ }$

$=\frac{ \sqrt{15 \cdot 5}$+ $\sqrt{15 \cdot 3}$}{5-3 }$

$=\frac{ \sqrt{75}$+ $\sqrt{45}$}{2 }$

$=\frac{ \sqrt{25}$ \cdot \sqrt {3}+ $\sqrt{9}$ \cdot $\sqrt{5}$}{2 }$

$=\frac{ 5\sqrt {3}$+3 $\sqrt{5}$}{2 }$

Divide: $$\frac{\sqrt{64}$}{$\sqrt{8}$}$

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The values are of the same power. Divide the terms.

$$\frac{\sqrt{64}$}{$\sqrt{8}$}= $\sqrt{\frac{64}{8}$}= $\sqrt{8}$ = $\sqrt{4}$\times $\sqrt{2}$= 2\sqrt2$

The correct answer is: $2\sqrt2$

Divide:

$18\textrm{ lbs }2\textrm{ oz } \div 5$

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We can write 18 pounds, 2 ounces as just ounces as follows:

$18\textrm{ lbs }2\textrm{ oz } = (18 \times 16 + 2) \textrm{ oz } = 290 \textrm{ oz }$

Divide:

$290 \div 5 = 58$

Convert 58 ounces to pounds and ounces as follows:

$58 \div 16 = 3 \textrm{ R }10$

That is, 3 pounds 10 ounces.

Simplify by rationalizing the denominator:

$$\frac{16}{4-\sqrt{6}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $4+$\sqrt{6}$$ :

$$\frac{16}{4-\sqrt{6}$}= $\frac{16(4+\sqrt{6}$)}{(4-$\sqrt{6}$) (4+$\sqrt{6}$)}$

$= $$\frac{16(4+\sqrt{6}$)}{4^{2}$$-($\sqrt{6}$)^{2}$ }$

$= $\frac{16(4+\sqrt{6}$)}{16-6}$

$= $\frac{16(4+\sqrt{6}$)}{10}$

$= $\frac{8(4+\sqrt{6}$)}{5}$

$= $\frac{ 32+8\sqrt{6}$}{5}$

Simplify by rationalizing the denominator:

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

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Multiply both numerator and denominator by the conjugate of the denominator, which is $6+$\sqrt{3}$$ :

$$\frac{\sqrt{6}$}{6-$\sqrt{3}$}$

$= $\frac{\sqrt{6}$(6+$\sqrt{3}$)}{(6-$\sqrt{3}$)(6+$\sqrt{3}$)}$

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

$= $\frac{6 \sqrt{6}$+ $\sqrt{18}$}{36-3 }$

$= $\frac{6 \sqrt{6}$+ $\sqrt{9}$ \cdot $\sqrt{2}$}{3 3 }$

$= $\frac{6 \sqrt{6}$+3$\sqrt{2}$}{3 3 }$

$= $\frac{2 \sqrt{6}$+$\sqrt{2}$}{11 }$

$= $\frac{\sqrt{2}$+ 2 $\sqrt{6}$}{11 }$

Simplify by rationalizing the denominator:

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

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Multiply both numerator and denominator by $\sqrt{6}$ :

$$\frac{3\sqrt{2}$+2$\sqrt{3}$}{$\sqrt{6}$}$

$= $\frac{(3\sqrt{2}$+2$\sqrt{3}$)\cdot $\sqrt{6}$}{$\sqrt{6}$\cdot $\sqrt{6}$}$

$= $\frac{3\sqrt{2}$\cdot $\sqrt{6}$+2$\sqrt{3}$\cdot $\sqrt{6}$}{6}$

$= $\frac{3\sqrt{12}$ +2$\sqrt{18}$ }{6}$

$= $\frac{3\cdot \sqrt{4}$ \cdot $\sqrt{3}$+2\cdot $\sqrt{9}$\cdot $\sqrt{2}$ }{6}$

$= $\frac{3\cdot 2 \cdot \sqrt{3}$+2\cdot3 \cdot $\sqrt{2}$ }{6}$

$= $\frac{6 \cdot \sqrt{3}$+6 \cdot $\sqrt{2}$ }{6}$

$= $\sqrt{3}$+ $\sqrt{2}$$