Simplifying Radical Expressions: Additional Practice (Set 2)

The index is odd, so the negative radicand is allowed and no absolute values will appear. Pull out perfect cubes: $-16 = (-8)(2)$ and $a^7 = a^6 \cdot a$, where $a^6 = (a^2)^3$.

$\sqrt[3]{-16a^7} = \sqrt[3]{(-8)(a^2)^3}\;\sqrt[3]{2a} = -2a^2\sqrt[3]{2a}$

Check by cubing: $(-2a^2)^3 = -8a^6$, and $-8a^6 \cdot 2a = -16a^7$, which is the radicand.

Every radicand is negative, so convert each factor to $i$ form before multiplying anything.

$\sqrt{-16} = 4i, \qquad \sqrt{-1} = i, \qquad \sqrt{-9} = 3i$

Multiply the three, collecting the powers of $i$:

$(4i)(i)(3i) = 12i^3 = 12 \cdot i^2 \cdot i = 12(-1)i = -12i$

Combining the radicands first would give $\sqrt{(-16)(-1)(-9)} = \sqrt{-144} = 12i$, the wrong answer, because the product rule does not survive negative radicands.

Start from the identity that always holds, $\sqrt{x^2} = |x|$. The question is when $|x| = -x$.

$|x| = -x \iff x \le 0$

If $x < 0$, then $|x| = -x$ by the definition of absolute value. If $x = 0$, then $|0| = 0 = -0$, so the endpoint works too and must be included. If $x > 0$, then $|x| = x$, which is positive, while $-x$ is negative, so the equation fails.

So the solution set is every $x \le 0$. The set $x < 0$ is not the answer: every one of its members does satisfy the equation, but it leaves out $x = 0$, so it is not the complete set.

This is the mirror image of $\sqrt{x^2} = x$, which holds exactly when $x \ge 0$.

Take the root of each perfect-square factor: $9 = 3^2$ and $x^4 = (x^2)^2$.

$\sqrt{9x^4} = \sqrt{3^2}\,\sqrt{(x^2)^2} = 3\,|x^2| = 3x^2$

The hypothesis $x < 0$ is a distraction here, and that is the point of the question. The exponent that comes out is $2$, which is even, so $x^2 \ge 0$ no matter what sign $x$ has, and no bars and no minus sign are needed. Test $x = -2$: the original is $\sqrt{9 \cdot 16} = 12$, and $3x^2 = 12$ agrees, while $-3x^2 = -12$ is negative and cannot be a square root.

Divide the exponent by the index: $12 \div 4 = 3$, so the candidate is $x^3$. Now check the sign, because the index is even and an even root is never negative.

$\sqrt[4]{x^{12}} = \sqrt[4]{(x^3)^4} = |x^3| = |x|^3$

The exponent that comes out is $3$, which is odd, so $x^3$ can be negative and the bars are mandatory. Check at $x = -2$: the original is $\sqrt[4]{4096} = 8$, and $|-2|^3 = 8$ agrees, while $x^3 = -8$ does not.

Simplify the radicand first: an even power of a negative number is positive.

$(-2)^6 = 64 \quad\Longrightarrow\quad \sqrt{(-2)^6} = \sqrt{64} = 8$

The identity gives the same thing directly: $\sqrt{x^6} = |x|^3$, and $|-2|^3 = 8$. The trap answer $-8$ comes from writing $\sqrt{x^6} = x^3 = (-2)^3$, which forgets that a square root cannot be negative.

The radicand is negative, so use the chapter-5 convention $\sqrt{-c} = i\sqrt{c}$ and then square.

$\left(\sqrt{-3}\right)^2 = \left(i\sqrt3\right)^2 = i^2\left(\sqrt3\right)^2 = (-1)(3) = -3$

Squaring a square root returns the radicand, and here the radicand is $-3$. Notice this does not contradict "a square root is never negative": that statement is about real principal roots, and $\sqrt{-3} = i\sqrt3$ is not a real number at all.

Replace the radical with what it actually equals, and the problem turns into an absolute-value equation.

$\sqrt{x^2} = |x| \quad\Longrightarrow\quad |x| = 7$

An absolute value of $7$ means a distance of $7$ from zero, in either direction:

$x = 7 \quad\text{or}\quad x = -7$

Both check out: $\sqrt{7^2} = 7$ and $\sqrt{(-7)^2} = \sqrt{49} = 7$. Note the difference from evaluating the symbol: $\sqrt{49}$ is a single number, but this equation has two solutions, and the absolute value is exactly what produces the second one.

The index is odd, so a negative radicand is fine and the root keeps the sign. Split the cube root over the quotient.

$\sqrt[3]{-\frac{8}{27}} = \frac{\sqrt[3]{-8}}{\sqrt[3]{27}} = \frac{-2}{3} = -\frac{2}{3}$

Check by cubing: $\left(-\tfrac23\right)^3 = -\tfrac{8}{27}$. Only an even root would reject this radicand.

Length is area divided by width, so divide the two radicals. Both radicands are positive, so the quotient rule lets you combine them under one sign.

$\text{length} = \frac{\sqrt{200}}{\sqrt{2}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10$

Check by multiplying back: $\sqrt2 \cdot 10 = 10\sqrt2 = \sqrt{200}$, the area. Subtracting the radicands to get $\sqrt{198}$ is not a rule and gives the wrong answer.

The radicand is a perfect-square trinomial. Recognise it before doing anything else.

$4x^2 - 12x + 9 = (2x - 3)^2$

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

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

The bars are required, because $2x - 3$ is negative whenever $x < \tfrac32$, and a square root never is. Check at $x = 0$: the original is $\sqrt{9} = 3$, and $|{-3}| = 3$ agrees, while $2x - 3 = -3$ does not.

Split off the largest perfect square inside, using $x^3 = x^2 \cdot x$.

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

The bars drop because the problem gives $x \ge 0$ (and the radicand $x^3$ would not be non-negative otherwise). Rational exponents confirm it: $\sqrt{x^3} = x^{3/2} = x^{1} \cdot x^{1/2} = x\sqrt{x}$. The distractors $\sqrt[3]{x^2}$ and $x^{2/3}$ are the same as each other, and both invert the index and the power.