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Solve the following division problems: 2√7, 5√9, 6√50, 7√24, 4√39, 3√19.
Mathematics
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14 stepsHere are the solutions to the problems:
- We need to simplify .
Identify if the number under the square root has any perfect square factors.
The number 7 is a prime number, so it does not have any perfect square factors other than 1.Since cannot be simplified further, the expression is already in its simplest form.
$$ 2\sqrt{7} $$ The simplified form is $\boxed{2\sqrt{7}}$.2. We need to simplify .
Evaluate the square root.
$$ \sqrt{9} = 3 $$Multiply the result by the coefficient.
$$ 5 \times 3 = 15 $$ The simplified form is $\boxed{15}$.3. We need to simplify .
Find the largest perfect square factor of 50.
$$ 50 = 25 \times 2 $$Rewrite the square root using the factors.
$$ \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} $$Multiply the simplified square root by the coefficient.
$$ 6 \times 5\sqrt{2} = 30\sqrt{2} $$ The simplified form is $\boxed{30\sqrt{2}}$.4. We need to simplify .
Find the largest perfect square factor of 24.
$$ 24 = 4 \times 6 $$Rewrite the square root using the factors.
$$ \sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6} $$Multiply the simplified square root by the coefficient.
$$ 7 \times 2\sqrt{6} = 14\sqrt{6} $$ The simplified form is $\boxed{14\sqrt{6}}$.5. We need to simplify .
Identify if the number under the square root has any perfect square factors.
The number 39 can be factored as $3 \times 13$. Neither 3 nor 13 are perfect squares, so 39 does not have any perfect square factors other than 1.Since cannot be simplified further, the expression is already in its simplest form.
$$ 4\sqrt{39} $$ The simplified form is $\boxed{4\sqrt{39}}$.6. We need to simplify .
Identify if the number under the square root has any perfect square factors.
The number 19 is a prime number, so it does not have any perfect square factors other than 1.Since cannot be simplified further, the expression is already in its simplest form.
$$ 3\sqrt{19} $$ The simplified form is $\boxed{3\sqrt{19}}$.
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