# Question #221ff

Feb 23, 2017

$\sqrt{72 {a}^{19}} = 6 \left\mid {a}^{9} \right\mid \sqrt{2 a}$

#### Explanation:

To find the square root of the given algebraic expression we
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should find the prime factorization.
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$72 = 2 \times 36$
$72 = 2 \times 2 \times 18$
$72 = 2 \times 2 \times 2 \times 9$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$72 = {2}^{3} \times {3}^{2}$
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$72 {a}^{19} = {2}^{3} \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a$
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$\sqrt{72 {a}^{19}} = \sqrt{{2}^{3} \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a}$
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$\sqrt{72 {a}^{19}} = \sqrt{{2}^{2} \times 2 \times {3}^{2} \times {\left({a}^{9}\right)}^{2} \times a}$
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$\sqrt{72 {a}^{19}} = \sqrt{{\left(2 \times 3 \times {a}^{9}\right)}^{2} \times 2 \times a}$
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$\sqrt{72 {a}^{19}} = \left\mid 2 \times 3 \times {a}^{9} \right\mid \sqrt{2 \times a}$
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$\sqrt{72 {a}^{19}} = 6 \left\mid {a}^{9} \right\mid \sqrt{2 a}$