Using Radicals with Elementary Operations

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Math › Using Radicals with Elementary Operations

Questions 1 - 3

1

Simplify:

$\sqrt{75}-\sqrt{27}$

$2\sqrt{3}$

$3\sqrt{5}$

$5\sqrt{2}$

$5\sqrt{3}$

$3\sqrt{2}$

Explanation

Try to group factors in pairs to get perfect squares under the square root:

$\sqrt{75}-\sqrt{27}= \sqrt{25\cdot 3}-\sqrt{9\cdot 3}=5\sqrt{3}-3\sqrt{3}=2\sqrt{3}$

2

What is the value of $5\sqrt5 + 3 - 4\sqrt{2} +9 - 3\sqrt{5}$ ?

$2\sqrt{5} - 4\sqrt{2} + 12$

$-2\sqrt{3} + 12$

$2\sqrt{10} - 4\sqrt{2} + 12$

$4\sqrt{7} - 12$

$2\sqrt{7} - 4\sqrt{7} + 12$

Explanation

When combining terms involving radicals, we can only combine the ones that have the same radical. For this problem, that means $4\sqrt2$ has to stay on its own while we can combine $5\sqrt5$ and $-3\sqrt5$ into $2\sqrt5$ . The simple integers can be combined too, giving us our answer with three seperate terms.

3

Simplify the expression:.

$\frac{x^{\frac{5}{6}}}{x^{\frac{7}{12}}}$

$\sqrt[4]{x}$

$\sqrt{x^6}$

$\sqrt{x^4}$

$\sqrt[6]{x}$

$\sqrt[6]{x^2}$

Explanation

Exponents in the denominator can be subtracted from exponents in the numerator.

$\small \frac{x^{\frac{5}{6}}}{x^{\frac{7}{12}}}=x^{\frac{5}{6}-\frac{7}{12}}=x^{\frac{10}{12}-\frac{7}{12}}=x^{\frac{3}{12}}=x^{\frac{1}{4}}$

Recall that $x^\frac{a}{b}=\sqrt[b]{x^a}$ .

Therefore, $x^\frac{1}{4}=\sqrt[4]{x}$ .

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