Fractional Exponents and Radicals: Free Response

The denominator of the exponent is the index of the root and the numerator is the power, so rooting first keeps the bases small.

For $64^{2/3}$ the index is $3$, and $4^{3} = 64$:

$64^{2/3} = \left(\sqrt[3]{64}\right)^{2} = 4^{2} = 16.$

For $243^{3/5}$ the index is $5$, and $3^{5} = 243$:

$243^{3/5} = \left(\sqrt[5]{243}\right)^{3} = 3^{3} = 27.$

The other order agrees, as it has to: $64^{2} = 4096$ and $\sqrt[3]{4096} = 16$. It simply asks you to take the cube root of a four-digit number instead of a two-digit one.

Translate one feature at a time. The index becomes a denominator, and a reciprocal becomes a minus sign.

The fourth root of $m^{3}$ has index $4$ and inside power $3$, so it is $m^{3/4}$, and the reciprocal negates that exponent:

$\frac{1}{\sqrt[4]{m^{3}}} = \frac{1}{m^{3/4}} = m^{-3/4}.$

Running the dictionary the other way, the minus sign in $t^{-5/6}$ asks for a reciprocal, and the $\tfrac{5}{6}$ that is left is a sixth root with the power $5$:

$t^{-5/6} = \frac{1}{t^{5/6}} = \frac{1}{\sqrt[6]{t^{5}}}.$

The minus sign moved the power downstairs. It did not make either value negative, and with $m$ and $t$ positive both results are positive.

The rule is all that has been assumed, so everything has to come out of it. Raise $a^{1/5}$ to the fifth power and multiply the exponents:

$\left(a^{1/5}\right)^{5} = a^{\frac{1}{5}\cdot 5} = a^{1} = a.$

So whatever $a^{1/5}$ names, its fifth power is $a$. That is exactly the defining property of a fifth root of $a$, so the rule has narrowed the field to those.

For an odd index the field holds one number. Raising to an odd power keeps the sign and keeps the order, so two different real numbers never share a fifth power, and every real number has a real fifth root. One candidate survives, and it is forced:

$a^{1/5} = \sqrt[5]{a}.$

With index $4$ the first half is untouched, since $\left(a^{1/4}\right)^{4} = a$ by the same multiplication of exponents. The second half fails, and it fails differently depending on $a$. An even power destroys the sign, so for $a > 0$ both a positive and a negative number raise to $a$, and the field holds two. For $a = 0$ only $0$ does, so the field holds one and the argument survives at that single value. For $a < 0$ nothing real raises to $a$ at all, and the field is empty. Uniqueness therefore fails for every positive $a$ and existence fails for every negative one, which is why an even index arrives with a convention (take the non-negative root) and a restriction (the radicand must not be negative), while an odd index needs neither.

Split each radicand into the largest factor that is a perfect power for its index, times whatever is left over.

For $\sqrt{147}$ the index is $2$, so hunt for a perfect square. Since $147 = 49 \cdot 3$ and $49 = 7^{2}$:

$\sqrt{147} = \sqrt{49}\,\sqrt{3} = 7\sqrt{3}.$

For $\sqrt[3]{192}$ the index is $3$, so hunt for a perfect cube instead. Since $192 = 64 \cdot 3$ and $64 = 4^{3}$:

$\sqrt[3]{192} = \sqrt[3]{64}\,\sqrt[3]{3} = 4\sqrt[3]{3}.$

In both answers the leftover radicand is $3$, which has no square factor and no cube factor beyond $1$, so neither will come down any further.

None of the three radicands is a perfect square, so simplify each one and then look at what is left.

$\sqrt{108} = \sqrt{36}\,\sqrt{3} = 6\sqrt{3}, \qquad \sqrt{243} = \sqrt{81}\,\sqrt{3} = 9\sqrt{3}.$

$\sqrt{48} = \sqrt{16}\,\sqrt{3} = 4\sqrt{3}.$

All three now carry the same index and the same radicand, so they are like radicals and the shared $\sqrt{3}$ comes out as a common factor:

$6\sqrt{3} + 9\sqrt{3} - 4\sqrt{3} = (6 + 9 - 4)\sqrt{3} = 11\sqrt{3}.$

The subtraction is handled entirely by the coefficients. Combining like radicals never changes the radicand, which is what makes the last line a single step rather than a calculation.

Adding like radicals is not a new rule. It is the distributive law applied to a common factor that happens to be a root:

$6\sqrt{3} + 9\sqrt{3} = (6 + 9)\sqrt{3}.$

That line works only because the very same number $\sqrt{3}$ sits inside both terms, and two things decide whether it does.

The radicand must match. $\sqrt{3}$ and $\sqrt{5}$ are different numbers, and neither is the other multiplied by anything you could write as a coefficient, so a sum of them has no shared factor to lift out.

The index must match too. $\sqrt{5}$ and $\sqrt[3]{5}$ are also different numbers: the first squares to $5$ and the second cubes to $5$, so they cannot both be $5$ raised to the same power. In exponent form the mismatch is plain:

$\sqrt{5} + \sqrt[3]{5} = 5^{1/2} + 5^{1/3}.$

Those are two different powers of $5$, and no exponent law adds two powers of a base into one; the laws combine powers that are multiplied. Simplifying cannot rescue it either, because $5$ is prime: it has no perfect-square factor and no perfect-cube factor beyond $1$, so each radical is already in its simplest form.

The volume of a cube is its edge used as a factor three times, so the edge is whatever number cubes to the volume. That is the cube root, which is the one-third power:

$s = 343^{1/3} = \sqrt[3]{343}.$

Since $7^{3} = 343$,

$s = 7 \text{ cm}.$

Check it forwards: $7 \times 7 \times 7 = 343$. The unit follows the same route, since a root of cubic centimetres returns plain centimetres, which is what a length has to be measured in.

The two quantities do not speak to each other directly, so go through the edge, which both of them are built from. A cube of edge $s$ has six square faces:

$S = 6s^{2}, \qquad V = s^{3}.$

The second equation says the edge is the cube root of the volume, that is $s = V^{1/3}$. Put that into the first and multiply the exponents:

$S = 6\left(V^{1/3}\right)^{2} = 6V^{2/3}.$

The exponent $\tfrac{2}{3}$ carries the whole story. The denominator $3$ undoes the cubing that produced the volume, and the numerator $2$ redoes the squaring that produces a face.

Test it on a cube of edge $7$. Through the formula, $6 \cdot 343^{2/3} = 6 \cdot 7^{2} = 294$; by counting faces, $6 \times 49 = 294$ square centimetres. The two routes agree.

Put $8V$ into the formula in place of $V$, and split the power across the product, which is what the product rule for roots licenses:

$6(8V)^{2/3} = 6 \cdot 8^{2/3} \cdot V^{2/3}.$

The new factor is $8^{2/3}$, and the exponent says cube root first, then square:

$8^{2/3} = \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4.$

So the second cube has surface area $4 \cdot 6V^{2/3}$, four times the first, and no particular volume ever had to be named for that to be true.

It is worth seeing why the answer is $4$ and not $8$. The cube root turns the volume factor $8$ into an edge factor of $2$, since eight times the volume means twice the edge, and the square then turns that edge factor into an area factor of $4$. Lengths scale once, areas twice, volumes three times, and the exponent $\tfrac{2}{3}$ is exactly two of those against three.

Take the student's sentences one at a time.

'The radicands $112$ and $175$ are different numbers.' True as written.

'Unlike radicals do not combine.' Also true, and it is exactly what the lesson says.

The break is the step between them: treating a difference in the radicands as first written as proof that the radicals are unlike. A radical that has not been simplified is not yet showing what number it is, and both of these have a perfect square hidden inside:

$\sqrt{112} = \sqrt{16 \cdot 7} = 4\sqrt{7}, \qquad \sqrt{175} = \sqrt{25 \cdot 7} = 5\sqrt{7}.$

Simplified, they share an index and a radicand after all, so they are like radicals and combine:

$4\sqrt{7} + 5\sqrt{7} = (4 + 5)\sqrt{7} = 9\sqrt{7}.$

The rule the student quoted was never in question. What went wrong is that the test was run on the wrong form of the terms, which is reading the packaging rather than the number.

Simplify every root first, then look at what you have.

For the first sum, $162 = 81 \cdot 2$ and $288 = 144 \cdot 2$:

$\sqrt{162} = 9\sqrt{2}, \qquad \sqrt{288} = 12\sqrt{2}.$

Same index, same radicand, so they combine:

$9\sqrt{2} + 12\sqrt{2} = 21\sqrt{2}.$

For the second sum, $242 = 121 \cdot 2$ and $150 = 25 \cdot 6$:

$\sqrt{242} = 11\sqrt{2}, \qquad \sqrt{150} = 5\sqrt{6}.$

The radicands are now $2$ and $6$, and neither has a square factor left to remove, so nothing further can bring them together:

$\sqrt{242} + \sqrt{150} = 11\sqrt{2} + 5\sqrt{6}.$

The two sums looked alike on the page. Only the simplified forms told them apart, which is the whole reason that step comes before any judgement.

Bracket each root between whole numbers by squaring, which needs no calculator.

Since $2^{2} = 4$ and $4 < 5$, we get $\sqrt{5} > 2$. Since $1^{2} = 1$ and $1 < 2$, we get $\sqrt{2} > 1$. Adding the two,

$\sqrt{5} + \sqrt{2} > 2 + 1 = 3.$

On the other side, $3^{2} = 9$ and $9 > 7$, so

$\sqrt{7} < 3.$

One number is above $3$ and the other is below it, so they are not the same number and the claim fails. Roots split across products, never across sums, and adding the radicands is a step no rule ever licensed.

The contrast with $6\sqrt{5} + 2\sqrt{5}$ is worth naming. There the very same number $\sqrt{5}$ sits in both terms, so it is a common factor and the distributive law lifts it out:

$6\sqrt{5} + 2\sqrt{5} = (6 + 2)\sqrt{5} = 8\sqrt{5}.$

Nothing of that kind is available for $\sqrt{5} + \sqrt{2}$. Neither term is the other multiplied by anything you could write as a coefficient, so there is nothing to lift out and nothing to collect, and the sum is already as simple as it will ever be.

Finish the inside completely, then take the root. Do not let the root and the power cancel on sight.

The square of $-9$ is positive, and a square root names the non-negative number that squares to its radicand:

$\sqrt{(-9)^{2}} = \sqrt{81} = 9.$

The sixth power of $-2$ is positive too, since six negative factors pair off completely, and the sixth root of $64$ is $2$:

$\sqrt[6]{(-2)^{6}} = \sqrt[6]{64} = 2.$

The cube of $-5$ keeps its sign, because three negative factors leave one unpaired, and an odd root of a negative number is negative:

$\sqrt[3]{(-5)^{3}} = \sqrt[3]{-125} = -5.$

Two of the three came back as something other than the number they started from. Which two, and why exactly those, is what the rest of the question settles.

Take the two parities separately, since part A has already shown that they behave differently.

Even $n$. Whatever $a$ is, $a^{n}$ is not negative, because the negative factors pair off. An even root of a number that is not negative is by convention the non-negative one, so the left-hand side is never negative. That alone rules out the unrestricted rule: an expression that is never negative cannot equal $a$ when $a$ is negative. Both $a$ and $-a$ raise to $a^{n}$, since an even power cannot tell them apart, and the one of those two that is not negative is $|a|$, so

$\sqrt[n]{a^{n}} = |a| \quad (n \text{ even}).$

Read back, that returns $a$ when $a \ge 0$ and $-a$ when $a < 0$, which is exactly what part A's first two values did.

Odd $n$. Now $a^{n}$ keeps the sign of $a$, and a real number has exactly one real $n$th root when $n$ is odd, so there is nothing to choose between. The number $a$ raises to $a^{n}$, so $a$ is that unique root:

$\sqrt[n]{a^{n}} = a \quad (n \text{ odd}).$

No restriction is needed and no absolute value appears, and that is the whole of the difference.

One fact drives all of it: an even power destroys the sign and an odd power preserves it. In an even power the negative factors pair off and leave no minus behind, while an odd power always has one factor left over to carry the sign.

Follow the consequences for an even index $n$. Raising to the $n$th power sends $a$ and $-a$ to the same place:

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

So a positive number has two real $n$th roots, differing only in sign. A symbol is allowed to name one number, so a choice has to be made, and the choice made is the non-negative root. The same collapse leaves a gap on the other side: nothing real has a negative even power, so a negative radicand has no even root at all, and $\sqrt[6]{-64}$ names nothing real. That is the restriction, and it is the same fact read backwards.

For an odd index neither consequence appears. Distinct numbers keep distinct odd powers, sign included:

$4^{3} = 64 \quad\text{and}\quad (-4)^{3} = -64.$

Each real number is therefore the odd power of exactly one real number. There is one root to name, so no convention is needed, and every real number, negative ones included, is somebody's odd power, so no radicand is barred.

The convention and the restriction are not two separate awkwardnesses, then. They are one fact about even powers, read once forwards and once backwards.