Simplifying Radical Expressions: Practice

Settle the domain first. The factor $y^2$ is never negative, so the sign of the radicand is decided by $x^3$: provided $y \ne 0$, the expression is real exactly when $x \ge 0$. (If $y = 0$ the radicand is $0$ whatever $x$ is, a degenerate case with nothing to simplify, so assume $y \ne 0$.) Note that $y$ itself is not restricted in sign: it may be negative.

Now extract the perfect squares, using $72 = 36 \cdot 2$ and $x^3 = x^2 \cdot x$.

$\sqrt{72x^3y^2} = \sqrt{36}\,\sqrt{x^2}\,\sqrt{y^2}\,\sqrt{2x} = 6\,|x|\,|y|\,\sqrt{2x}$

The domain gives $x \ge 0$, so $|x| = x$ and those bars come off. Nothing pins down the sign of $y$, so its bars must stay.

$\sqrt{72x^3y^2} = 6x|y|\sqrt{2x}$

Test $x = 2$, $y = -1$: the original is $\sqrt{576} = 24$, and $6(2)|-1|\sqrt{4} = 24$ agrees, while the bar-free $6xy\sqrt{2x}$ would give $-24$, a value no square root can take.

Test the cases. If $a$ and $b$ are both non-negative, the rule is a theorem and always holds. If exactly one is negative, say $a = -m$ with $m > 0$ and $b \ge 0$, then $\sqrt{a}\sqrt{b} = i\sqrt{m}\sqrt{b} = i\sqrt{mb}$, and $\sqrt{ab} = \sqrt{-mb} = i\sqrt{mb}$, so the two sides agree.

If both are negative, $a = -m$ and $b = -n$ with $m, n > 0$, the two sides differ by a sign:

$\sqrt{a}\,\sqrt{b} = (i\sqrt m)(i\sqrt n) = -\sqrt{mn}, \qquad \sqrt{ab} = \sqrt{mn}$

So the failure happens exactly in the both-negative case. "Whenever $a$ is negative" is too strong ($a = -4$, $b = 9$ works fine), and "$ab < 0$" is wrong in both directions: that same $a = -4, b = 9$ has $ab < 0$ yet the rule holds, while $a = b = -1$ has $ab > 0$ and the rule fails.

Both radicands are positive because $x > 0$, so the quotient rule applies and the two radicals can be combined into one.

$\frac{\sqrt{54x^3}}{\sqrt{6x}} = \sqrt{\frac{54x^3}{6x}} = \sqrt{9x^2}$

Now take the root:

$\sqrt{9x^2} = 3|x| = 3x$

The bars come off because the problem gives $x > 0$. Dividing inside the radical first is what makes this quick: the messy radicands cancel into a perfect square.

Different indices cannot be combined directly, so convert both radicals to rational exponents, where the exponent rules do the work.

$\sqrt{x}\cdot\sqrt[3]{x} = x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3}$

Add the fractions over the common denominator $6$:

$\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \quad\Longrightarrow\quad x^{5/6} = \sqrt[6]{x^5}$

The index is the denominator ($6$) and the power is the numerator ($5$). Since $5$ and $6$ share no common factor, this is already in simplest form.

Two negative radicands, so convert to $i$ form before multiplying. Never combine them under one radical.

$\sqrt{-5}\cdot\sqrt{-5} = \left(i\sqrt5\right)\left(i\sqrt5\right) = i^2\left(\sqrt5\right)^2 = (-1)(5) = -5$

The trap answer is $5$, from writing $\sqrt{(-5)(-5)} = \sqrt{25} = 5$. The product rule needs non-negative radicands; here it lands on the wrong square root of $25$, missing by a factor of $-1$.

The identity is $\sqrt{u^2} = |u|$, so with $u = x - 4$,

$\sqrt{(x-4)^2} = |x - 4|.$

Now resolve the absolute value using the hypothesis. Since $x < 4$, the quantity $x - 4$ is negative, and the absolute value of a negative number is its opposite.

$|x - 4| = -(x - 4) = 4 - x$

Sanity check at $x = 1$: the original is $\sqrt{(-3)^2} = 3$, and $4 - x = 3$ agrees, while $x - 4 = -3$ has the wrong sign.

The radicand is a product of two non-negative factors, so split it and take each root.

$\sqrt{25(x+1)^2} = \sqrt{25}\,\sqrt{(x+1)^2} = 5\,|x + 1|$

The bars must survive: $x$ is unrestricted, so $x + 1$ can be negative (take $x = -3$), while the square root cannot. Test $x = -3$: the original is $\sqrt{25 \cdot 4} = 10$, and $5|-2| = 10$ agrees, while $5(x+1) = -10$ does not. Note also that the bars go around the whole binomial, not just the $x$.

Use the identity, then resolve the absolute value with the hypothesis $x < 0$.

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

Because $x$ is negative, $|x| = -x$, so the two terms cancel:

$(-x) + x = 0$

Check at $x = -7$: $\sqrt{49} + (-7) = 7 - 7 = 0$. The answer $2x$ would come from writing $\sqrt{x^2} = x$, which is precisely what fails for negative $x$.

The index is $5$, which is odd, so a negative radicand is allowed and no bars are needed. Write the radicand as a perfect fifth power.

$-32x^{10} = (-2)^5\left(x^2\right)^5 = \left(-2x^2\right)^5$

The fifth root undoes the fifth power exactly:

$\sqrt[5]{-32x^{10}} = -2x^2$

Check: $(-2x^2)^5 = -32x^{10}$. Note the answer is negative for every $x \ne 0$, which is fine, because an odd root is allowed to be negative.

Since $\sqrt{x^2} = |x|$, the statement $\sqrt{x^2} = x$ says $|x| = x$, which holds exactly when $x \ge 0$. So it fails precisely for negative $x$.

$\sqrt{(-5)^2} = \sqrt{25} = 5 \ne -5$

The other three values are non-negative, so the statement is true for each of them (for instance $\sqrt{0^2} = 0$ and $\sqrt{25^2} = 25$).

Work from the inside out. The inner radical is a plain square root.

$\sqrt{64} = 8$

Now take the cube root of that:

$\sqrt[3]{8} = 2$

Rational exponents give the same answer in one line, and they explain why nesting the roots multiplies the indices: $\left(64^{1/2}\right)^{1/3} = 64^{1/6} = 2$, since $2^6 = 64$.

The identity gives $\sqrt{a^2} = |a|$ for every real $a$. Now use the hypothesis to unpack the absolute value.

$a < 0 \quad\Longrightarrow\quad |a| = -a$

So $\sqrt{a^2} = -a$. This looks negative but is not: $a$ is negative, so $-a$ is positive, exactly as a square root must be. Check at $a = -6$: $\sqrt{36} = 6$, and $-a = 6$.