Simplifying Radical Expressions: Practice

Find the largest perfect square dividing $200$. Since $200 = 100 \cdot 2$, that square is $100$.

$\sqrt{200} = \sqrt{100}\,\sqrt{2} = 10\sqrt{2}$

The perfect square leaves the radical as its root, $10$, not as $100$. Check: $(10\sqrt2)^2 = 100 \cdot 2 = 200$.

Both radicands are negative, so the product rule $\sqrt{ab} = \sqrt a \sqrt b$ does not apply. Convert each radical to $i$ form first, then multiply.

$\sqrt{-9}\cdot\sqrt{-16} = (3i)(4i) = 12i^2 = -12$

The trap answer is $12$, from writing $\sqrt{(-9)(-16)} = \sqrt{144} = 12$. That step misapplies the product rule to two negative numbers, and it lands on the wrong one of the two square roots of $144$.

The radicand $45x^2$ is non-negative for every real $x$, so $x$ is unrestricted and the answer must be right for negative $x$ too. Split off the perfect squares: $45 = 9 \cdot 5$.

$\sqrt{45x^2} = \sqrt{9}\,\sqrt{x^2}\,\sqrt{5} = 3\,|x|\,\sqrt{5}$

The bars are required: at $x = -1$ the original is $\sqrt{45} \approx 6.7$, a positive number, while $3x\sqrt5$ would be negative.

First the domain. The radicand $32x^5$ is non-negative only when $x \ge 0$, so that hypothesis comes free with the problem and no absolute values will be needed.

Split each factor into a perfect square times a remainder: $32 = 16 \cdot 2$ and $x^5 = x^4 \cdot x$.

$\sqrt{32x^5} = \sqrt{16}\,\sqrt{x^4}\,\sqrt{2x} = 4x^2\sqrt{2x}$

The leftover $2x$ stays inside, since neither factor is a perfect square.

The index is odd, so the negative radicand is fine and no absolute values appear. Write the radicand as a perfect cube.

$-64x^6 = (-4)^3 (x^2)^3 = (-4x^2)^3$

Taking the cube root undoes the cube exactly:

$\sqrt[3]{-64x^6} = -4x^2$

Check: $(-4x^2)^3 = -64x^6$, as required.

The index $4$ and the exponent hiding in $25 = 5^2$ share the factor $2$, so the index can be reduced. Convert to a rational exponent and reduce the fraction.

$\sqrt[4]{25} = \sqrt[4]{5^2} = 5^{2/4} = 5^{1/2} = \sqrt{5}$

Check it by raising to the fourth power: $(\sqrt5)^4 = (5^{1/2})^4 = 5^2 = 25$.

Write the radicand as a power of a prime: $27 = 3^3$. Now the index $6$ and the exponent $3$ share the factor $3$.

$\sqrt[6]{27} = \sqrt[6]{3^3} = 3^{3/6} = 3^{1/2} = \sqrt{3}$

Reducing $3/6$ to $1/2$ is exactly what cancelling the common factor means. Check: $(\sqrt3)^6 = 3^3 = 27$.

The radicand is non-negative for every real $x$, so $x$ is unrestricted. Split the root over the quotient, which is allowed since the numerator is $\ge 0$ and the denominator is $> 0$.

$\sqrt{\frac{x^2}{49}} = \frac{\sqrt{x^2}}{\sqrt{49}} = \frac{|x|}{7}$

The numerator needs the bars, because $\sqrt{x^2} = |x|$ and the whole expression must come out non-negative. The denominator is a plain $7$, since $\sqrt{49} = 7$.

By the definition of the principal square root, $\sqrt{7}$ is the non-negative number whose square is $7$. Squaring it therefore returns the radicand.

$\left(\sqrt{7}\right)^2 = 7$

This direction never needs an absolute value: for $a \ge 0$, $(\sqrt a)^2 = a$ exactly. It is the other order, $\sqrt{a^2}$, that needs the bars.

The two radicals behave differently, and that is the whole point. The even root must return a non-negative value, so it gives $|-6|$; the odd root keeps the sign, so it gives $-6$.

$\sqrt{(-6)^2} = \sqrt{36} = 6, \qquad \sqrt[3]{(-6)^3} = \sqrt[3]{-216} = -6$

Adding them:

$6 + (-6) = 0$

If you had written $\sqrt{(-6)^2} = -6$, you would have gotten $-12$, which is the classic error.

Write the radicand as a perfect fourth power: $16 = 2^4$ and $x^8 = (x^2)^4$.

$\sqrt[4]{16x^8} = \sqrt[4]{2^4}\,\sqrt[4]{(x^2)^4} = 2\,|x^2| = 2x^2$

The index is even, so an absolute value is called for, but $x^2 \ge 0$ for every real $x$, so $|x^2| = x^2$ and the bars are redundant. The exponent that comes out is even, which is exactly when bars can be dropped.

There is no rule that splits a radical over a sum, so combine the like terms inside first.

$9x^2 + 16x^2 = 25x^2$

Now take the root of the product:

$\sqrt{25x^2} = \sqrt{25}\,\sqrt{x^2} = 5\,|x|$

The wrong answer $7|x|$ comes from splitting the sum as $\sqrt{9x^2} + \sqrt{16x^2} = 3|x| + 4|x|$, which is not a legal step. The bars are needed because $x$ is unrestricted.