Simplifying Radical Expressions: Additional Practice (Set 1)

Check the domain first. The factor $x^4$ is never negative, so the sign of the radicand is decided by $y^3$: provided $x \ne 0$, the expression is real exactly when $y \ge 0$. (If $x = 0$ the radicand is $0$ whatever $y$ is, a degenerate case with nothing to simplify, so assume $x \ne 0$.) That hypothesis comes free with the problem, and it means no absolute values will survive.

Extract perfect squares, using $48 = 16 \cdot 3$ and $y^3 = y^2 \cdot y$.

$\sqrt{48x^4y^3} = \sqrt{16}\,\sqrt{x^4}\,\sqrt{y^2}\,\sqrt{3y} = 4x^2\,|y|\,\sqrt{3y}$

The $\sqrt{x^4} = x^2$ needs no bars (the exponent coming out is even), and $|y| = y$ because the domain forces $y \ge 0$:

$\sqrt{48x^4y^3} = 4x^2y\sqrt{3y}$

The index is odd, so the negative radicand is fine and no bars are needed. Pull out perfect cubes: $-24 = (-8)(3)$ and $a^{10} = a^9 \cdot a$, where $a^9 = (a^3)^3$.

$\sqrt[3]{-24a^{10}} = \sqrt[3]{(-8)(a^3)^3}\;\sqrt[3]{3a} = -2a^3\sqrt[3]{3a}$

Check by cubing: $(-2a^3)^3 = -8a^9$, and $-8a^9 \cdot 3a = -24a^{10}$, which is the radicand. Taking $a^{10} = a^9 \cdot a$ (rather than $a^6 \cdot a^4$) is what leaves the smallest possible power inside.

The radicand is a perfect-square trinomial, so factor it before touching the radical.

$x^2 + 6x + 9 = (x + 3)^2$

Now apply $\sqrt{u^2} = |u|$ with $u = x + 3$:

$\sqrt{(x+3)^2} = |x + 3|$

The bars are required because $x + 3$ is negative whenever $x < -3$, while a square root never is. Test $x = -5$: the original is $\sqrt{25 - 30 + 9} = \sqrt{4} = 2$, and $|-2| = 2$ agrees, while $x + 3 = -2$ does not. The bars go around the whole binomial, not just the $x$.

The radicand is non-negative for every allowed $x$ and $y$, so neither variable is restricted in sign. Split the root over the quotient.

$\sqrt{\frac{49x^6}{y^4}} = \frac{\sqrt{49}\,\sqrt{x^6}}{\sqrt{y^4}} = \frac{7\,|x|^3}{y^2}$

The two variables behave differently, and the exponent that comes out is what decides. From $x^6$ the exponent $3$ emerges, which is odd, so $x^3$ could be negative and the bars are mandatory. From $y^4$ the exponent $2$ emerges, which is even, so $y^2 \ge 0$ already and bars would be redundant.

A whole number is a perfect cube exactly when every prime in its factorization appears a number of times that is a multiple of $3$. Factor the radicand.

$54 = 2 \cdot 3^3$

The $3^3$ is already a perfect cube, but the prime $2$ appears once, so it needs two more copies of itself and nothing else.

$54 \cdot 4 = 216 = 6^3 \quad\Longrightarrow\quad \sqrt[3]{216} = 6$

So $n = 4$. Note this differs from the square-root version of the question: a cube root needs the primes in groups of three, not pairs, so $n = 2$ is not enough ($108$ is not a cube).

Different indices cannot be compared directly, so raise everything to a power that clears all of them at once. The indices are $3$, $4$, $2$, and $1$, so use the twelfth power (raising to a positive power preserves order among positive numbers).

$\left(\sqrt[3]{30}\right)^{12} = 30^4 = 810{,}000, \qquad \left(\sqrt[4]{80}\right)^{12} = 80^3 = 512{,}000$

$3^{12} = 531{,}441, \qquad \left(\sqrt{10}\right)^{12} = 10^6 = 1{,}000{,}000$

The largest twelfth power belongs to $\sqrt{10}$, so $\sqrt{10}$ is the largest number. The full order is $\sqrt{10} > \sqrt[3]{30} > 3 > \sqrt[4]{80}$.

Apply the identity $\sqrt{u^2} = |u|$ with $u = 2 - \sqrt5$, then decide the sign of $u$.

$\sqrt{\left(2 - \sqrt5\right)^2} = \left|2 - \sqrt5\right|$

Since $\sqrt5 \approx 2.236 > 2$, the quantity $2 - \sqrt5$ is negative, so its absolute value is its opposite.

$\left|2 - \sqrt5\right| = -(2 - \sqrt5) = \sqrt5 - 2$

The answer is about $0.236$, a positive number, as a square root must be. Writing $2 - \sqrt5$ would give a negative value, which the radical can never produce.

Write the radicand as a perfect fourth power: $16 = 2^4$, $x^{12} = (x^3)^4$, and $y^8 = (y^2)^4$.

$\sqrt[4]{16x^{12}y^8} = 2\,\left|x^3\right|\,\left|y^2\right|$

The index is even, so each factor comes out inside absolute values. Resolve them one at a time. The exponent $3$ from $x$ is odd, so $x^3$ can be negative and $|x^3| = |x|^3$ must keep its bars. The exponent $2$ from $y$ is even, so $y^2 \ge 0$ already and $|y^2| = y^2$.

$\sqrt[4]{16x^{12}y^8} = 2|x|^3y^2$

Both radicands are non-negative for every real $x$ (note $2x^2 \ge 0$), so the product rule applies and the radicals can be combined.

$\sqrt{50}\cdot\sqrt{2x^2} = \sqrt{100x^2}$

Now take the root of each factor:

$\sqrt{100x^2} = \sqrt{100}\,\sqrt{x^2} = 10\,|x|$

The bars stay, because nothing restricts the sign of $x$ while the square root is committed to being non-negative.

If $\sqrt{N} = 3\sqrt5$, then squaring both sides recovers $N$, since $3\sqrt5$ is non-negative.

$N = \left(3\sqrt5\right)^2 = 3^2\left(\sqrt5\right)^2 = 9 \cdot 5 = 45$

Check the simplification in the forward direction as well: $\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt5$. The common error is to square only one factor, giving $9 \cdot \sqrt5$ or $9 \cdot 5^2$.

Simplify the fraction inside the radical before doing anything else.

$\frac{54x^4}{2x} = 27x^3$

Now the radicand is a perfect cube, and the index is odd, so no absolute value is needed:

$\sqrt[3]{27x^3} = 3x$

The answer keeps the sign of $x$, exactly as an odd root should. Writing $3|x|$ would be wrong: at $x = -1$ the radicand is $-27$ and the cube root is $-3$, not $+3$.

Test each against the two conditions in play: no perfect $n$-th power factors left inside, and no common factor between the index and the exponents.

$\sqrt[4]{4} = \sqrt[4]{2^2} = 2^{2/4} = \sqrt2, \qquad \sqrt[6]{8} = \sqrt[6]{2^3} = 2^{3/6} = \sqrt2$

Both of those reduce, because the index shares a factor with the exponent. And $\sqrt{20} = 2\sqrt5$, because $20$ hides the perfect square $4$.

In $2\sqrt[3]{9}$ the radicand is $9 = 3^2$, which has no perfect cube factor, and the index $3$ shares no factor with the exponent $2$. Nothing can be extracted or reduced, so it is in simplest form.