Simplifying Radical Expressions: Practice

The radical sign asks for the principal square root: the one non-negative number whose square is $64$.

$8^2 = 64 \quad\text{and}\quad 8 \ge 0$

So $\sqrt{64} = 8$. It is true that $(-8)^2 = 64$ as well, but $-8$ is negative, so it is not what the symbol names. Halving $64$ gives $32$, which is a different operation entirely.

Look for the non-negative number that squares to $121$.

$11^2 = 121$

So $\sqrt{121} = 11$. The symbol returns a single value, so $\pm 11$ is not an answer to this question.

Squaring destroys sign, so a positive number is the square of two different reals.

$5^2 = 25 \quad\text{and}\quad (-5)^2 = 25$

The equation has two solutions, $x = 5$ and $x = -5$. Keep this apart from the value of the symbol $\sqrt{25}$, which is $5$ alone.

Work from the inside out. The radical gives the principal root, and the minus sign in front then negates it.

$-\sqrt{36} = -(6) = -6$

This is how the negative square root of $36$ is written: the symbol itself is always non-negative, so the sign is applied outside it.

The index is odd, so a negative radicand is allowed and the root carries the sign of the radicand.

$(-5)^3 = (-5)(-5)(-5) = -125$

So $\sqrt[3]{-125} = -5$. Odd roots of negative numbers are perfectly real; it is even roots that reject a negative radicand.

An even power of a real number is never negative, so an even root cannot accept a negative radicand over the reals.

$r^2 \ge 0 \ \text{ for every real } r, \ \text{ so no real } r \text{ has } r^2 = -4$

So $\sqrt{-4}$ is not real (it is the complex number $2i$). The other three are fine: $\sqrt{16} = 4$, and the odd roots $\sqrt[3]{-8} = -2$ and $\sqrt[5]{-32} = -2$ happily take negative radicands.

Find the largest perfect square that divides $50$. That is $25$, since $50 = 25 \cdot 2$.

$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}$

Only the $\sqrt{25}$ leaves the radical, as the number $5$; the $25$ itself does not come out. A quick check: $5\sqrt{2} \approx 7.07$, and $7.07^2 \approx 50$.

The radicand is a sum, so add before taking the root.

$\sqrt{9 + 16} = \sqrt{25} = 5$

The tempting wrong answer is $7$, which comes from splitting the radical over the sum as $\sqrt{9} + \sqrt{16} = 3 + 4$. There is no such rule: radicals distribute over products and quotients, never over sums.

The quotient rule splits a root over a fraction, since $16 \ge 0$ and $81 > 0$.

$\sqrt{\frac{16}{81}} = \frac{\sqrt{16}}{\sqrt{81}} = \frac{4}{9}$

Both the numerator and the denominator are perfect squares, so the radical disappears completely.

The largest perfect square dividing $75$ is $25$, since $75 = 25 \cdot 3$.

$\sqrt{75} = \sqrt{25}\,\sqrt{3} = 5\sqrt{3}$

The $3$ stays inside because it has no square factor. Note that $\sqrt{75}$ is not $15$: squaring $15$ gives $225$, not $75$.

Split off the largest perfect square factor: $12 = 4 \cdot 3$.

$\sqrt{12} = \sqrt{4}\,\sqrt{3} = 2\sqrt{3}$

Only the perfect square comes out, and it comes out as its root, $2$.

The small number in the crook of the radical sign is the index, and the number underneath is the radicand.

$\sqrt[4]{7}: \quad \text{index} = 4, \quad \text{radicand} = 7$

The index says which power to undo, so $\sqrt[4]{7}$ is the non-negative number whose fourth power is $7$. An index of $2$ is the square-root case, and it is the one index that is left unwritten.