How to divide square roots - SSAT Upper Level Quantitative (Math)

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Question

Divide:

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

Answer

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.

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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

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

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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

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

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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

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

This is not among the given choices.

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Question

Simplify by rationalizing the denominator:

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

Answer

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

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

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

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

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

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

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

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

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Question

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

Answer

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: $2\sqrt2$

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