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Radicals and Rational Exponents

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

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Press Play: the idea and a worked example, explained aloud (about 1:17).

Radicals and Rational Exponents

A root undoes a power.

The square root of 49 is 7.
The cube root of 64 is 4.

\begin{aligned} \sqrt{49} &= 7 \end{aligned}

\begin{aligned} \sqrt[3]{64} &= 4 \end{aligned}

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Read what is said

A root undoes a power. Radicals and fractional exponents show up on the ACT again and again, usually as simplify this or find the value questions. Most of the work is a few clean habits.
A fractional exponent is just a root written another way. The bottom of the fraction is the index of the root, and the top is the power. Keep that order straight.
To simplify a square root, look for the largest square factor. Here thirty six goes into seventy two, so it can come out of the root as a six. What is left under the root is a two.
Radicals only add when they are alike, and they look unlike until you simplify. Simplify each root first. Then all three have the same radicand, so just combine the numbers in front.
Now a rational power. Take the root first, because it keeps the numbers small. The bottom of the exponent is four, so take the fourth root of sixteen. Then use the top number, three, as the power.
Here is the trap the answer choices are built on. A root of a sum is not the sum of the roots. Add under the root first, and you get a root of twenty five. Do not split it into three plus four.
Remember: simplify each radical first, and read a fractional exponent as root on the bottom, power on top. Now practice it with new numbers.

The idea

A root undoes a power

The square root of 49, written $\sqrt{49}$ , is the positive number whose square is 49. The cube root of 64, written $\sqrt[3]{64}$ , is the number whose cube is 64. The small number on the root sign is the index; a square root has index 2, which is not written. The number under the sign is the radicand.

\begin{aligned} r &= \sqrt{49} \\ &= 7 \\ s &= 64^{\frac{1}{3}} \\ &= 4 \\ y &= s^{3} \\ &= 4^{3} \\ &= 64 \end{aligned}

A square shows the idea. Its area is the side squared, so its side is the square root of its area. Drag the slider: most areas do not give a whole-number side.

Which numbers have a real root

No real number squared is negative, so a negative number has no real square root: $\sqrt{-9}$ is not a real number. The same holds for every even index. An odd root is different:

\begin{aligned} c &= (-2)^{3} \\ &= -8 \\ r &= (-8)^{\frac{1}{3}} \\ &= -2 \end{aligned}

Watch the two graphs draw themselves. The square root starts at 0 and exists only to the right; the cube root runs through negative numbers too.

Rule.

An even root $\sqrt[n]{a}$ (index 2, 4, 6, ...) is a real number only when $a \ge 0$ . An odd root is real for every $a$ .

Simplifying a square root

Because $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a, b \ge 0$ , a square factor can be taken out of the root. For $\sqrt{72}$ , the largest square factor of 72 is 36:

\begin{aligned} k &= \sqrt{36} \\ &= 6 \\ R &= \sqrt{72} \\ &= 6\sqrt{2} \end{aligned}

So $\sqrt{72} = \sqrt{36}\sqrt{2} = {6\sqrt{2}}$ . The picture shows why: a square of area 72 is made of 36 small squares, each of area 2 and side $\sqrt{2}$ . Along each edge there are $6$ of them.

Square roots with the same radicand are like radicals, and they add as like terms do: $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$ . Simplify first; $\sqrt{12} + \sqrt{27}$ only looks unlike:

\begin{aligned} S &= \sqrt{12} + \sqrt{27} \\ &= 5\sqrt{3} \end{aligned}

Rationalizing a denominator

An answer is usually written with no root in the denominator. Multiply top and bottom by the root, which changes nothing because it multiplies by 1. For a bottom such as $3 - \sqrt{2}$ , multiply by $3 + \sqrt{2}$ : the product of the two is $9 - 2$ , with no root left.

\begin{aligned} A &= \frac{6}{\sqrt{3}} \\ &= 2\sqrt{3} \\ B &= \frac{7}{3 - \sqrt{2}} \\ &= 3 + \sqrt{2} \end{aligned}

Rational exponents

Multiply $x^{1/2}$ by itself and the exponents add to 1, so $x^{1/2}$ is $\sqrt{x}$ . In the same way $x^{1/3}$ is $\sqrt[3]{x}$ .

Rule.

$x^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m$ . The bottom of the fraction is the index of the root and the top is the power. A negative exponent still means 1 over the power.

Take the root first to keep the numbers small. For $8^{2/3}$ , take the cube root $r$ of 8, then square it:

\begin{aligned} r &= 8^{\frac{1}{3}} \\ &= 2 \\ y &= r^{2} \\ &= 2^{2} \\ &= 4 \end{aligned}

So $8^{2/3} = {4}$ .

Warning.

A root of a sum is not the sum of the roots: $\sqrt{9 + 16} = \sqrt{25} = 5$ , not $3 + 4$ . Other slips the ACT's choices are built on: leaving a square factor inside the root, swapping the index and the power ( $x^{2/3}$ is the cube root of $x^{2}$ ), and thinking an even root of a negative number is real.

Worked examples

Each example is one of the kinds of question the ACT asks about radicals and rational exponents. The numbers in the working are found by the engine.

Example.

1. A formula that gives a root. On a clear day, a person whose eyes are $h$ feet above the sea can see about $d = \sqrt{1.5h}$ miles to the horizon. How far can a person see from a tower 48 feet high? Give the answer as a simplified radical.

Put 48 in for $h$ . The number under the root is $1.5 \times 48$ ; then take out the largest square factor.

\begin{aligned} u &= 1.5h \\ &= 1.5 \times 48 \\ &= 72 \\ k &= \sqrt{36} \\ &= 6 \\ d &= \sqrt{u} \\ &= \sqrt{72} \\ &= 6\sqrt{2} \end{aligned}

$72$ is 36 times 2, so $\sqrt{ {72} } = \sqrt{36}\sqrt{2} = {6\sqrt{2}}$ miles, about $8.49$ miles. A choice such as $2\sqrt{18}$ has the same value but is not simplified: a square factor, 9, is still inside.

Example.

2. Adding radicals. Simplify $2\sqrt{50} + 3\sqrt{8} - \sqrt{18}$ .

The radicands look different, but each has a square factor. Simplify each root first; then they are like radicals and the numbers in front add.

\begin{aligned} a &= 2\sqrt{50} \\ &= 10\sqrt{2} \\ b &= 3\sqrt{8} \\ &= 6\sqrt{2} \\ c &= \sqrt{18} \\ &= 3\sqrt{2} \\ S &= a + b - c \\ &= 10\sqrt{2} + 6\sqrt{2} - 3\sqrt{2} \\ &= 13\sqrt{2} \end{aligned}

Adding the radicands, as if $\sqrt{50} + \sqrt{8}$ were $\sqrt{58}$ , is the usual wrong answer: a root of a sum is not a sum of roots.

Example.

3. Multiplying radicals. Simplify $(2\sqrt{3})(5\sqrt{6})$ and $(\sqrt{7} + 2)(\sqrt{7} - 2)$ .

Multiply the numbers outside and the numbers inside, then simplify the root:

\begin{aligned} a &= 2 \times 5 \\ &= 10 \\ b &= 3 \times 6 \\ &= 18 \\ P &= 2\sqrt{3} \times 5\sqrt{6} \\ &= 30\sqrt{2} \end{aligned}

So it is ${10}\sqrt{ {18} }$ , and $18$ has the square factor 9. In the second product the middle terms cancel, as in $(p + q)(p - q) = p^2 - q^2$ :

\begin{aligned} Q &= (\sqrt{7} + 2) \times (\sqrt{7} - 2) \\ &= 3 \end{aligned}

Two numbers like $\sqrt{7} + 2$ and $\sqrt{7} - 2$ are conjugates: their product has no root in it. The next example uses that.

Example.

4. Rationalizing a denominator. Write $\frac{10}{\sqrt{5}}$ and $\frac{4}{3 - \sqrt{5}}$ with no root in the denominator.

For the first, multiply top and bottom by $\sqrt{5}$ . For the second, multiply top and bottom by the conjugate $3 + \sqrt{5}$ . Work the new top $T$ and bottom $B$ :

\begin{aligned} A &= \frac{10}{\sqrt{5}} \\ &= 2\sqrt{5} \\ T &= 4 \times (3 + \sqrt{5}) \\ &= 12 + 4\sqrt{5} \\ B &= (3 - \sqrt{5}) \times (3 + \sqrt{5}) \\ &= 4 \\ C &= \frac{4}{3 - \sqrt{5}} \\ &= 3 + \sqrt{5} \end{aligned}

The bottom became $4$ , a whole number, and dividing the top by it gives $3 + \sqrt{5}$ . Multiplying top and bottom by $3 - \sqrt{5}$ in place of the conjugate leaves a root in the bottom.

Example.

5. The value of a rational power. Find $16^{3/4}$ and $27^{-2/3}$ .

The bottom of the exponent is the index of the root and the top is the power. Take the root first. For $16^{3/4}$ , the fourth root of 16, then its cube:

\begin{aligned} r &= 16^{\frac{1}{4}} \\ &= 2 \\ y &= r^{3} \\ &= 2^{3} \\ &= 8 \end{aligned}

For $27^{-2/3}$ , the minus sign means 1 over the power. Take the cube root of 27, square it, then take 1 over the result:

\begin{aligned} s &= 27^{\frac{1}{3}} \\ &= 3 \\ v &= \frac{1}{s^{2}} \\ &= \frac{1}{3^{2}} \\ &= \frac{1}{9} \end{aligned}

The engine works both powers in one step each and agrees:

\begin{aligned} Y &= 16^{\frac{3}{4}} \\ &= 8 \\ V &= 27^{-\frac{2}{3}} \\ &= \frac{1}{9} \end{aligned}

Neither answer is negative. Multiplying 27 by the exponent is the slip to avoid: an exponent is not a factor.

Example.

6. Radicals with a variable. For positive $x$ , write $\sqrt{x} \cdot \sqrt[3]{x^2}$ as one power of $x$ .

Write each radical as a power: $\sqrt{x} = x^{1/2}$ and $\sqrt[3]{x^2} = x^{2/3}$ . The powers are multiplied, so the exponents add:

\begin{aligned} e &= \frac{1}{2} + \frac{2}{3} \\ &= \frac{7}{6} \\ P &= x^{\frac{1}{2}}x^{\frac{2}{3}} \\ &= x^{\frac{7}{6}} \end{aligned}

So the product is $x^{\frac{7}{6}}$ , the sixth root of $x^{7}$ . Adding tops and bottoms, $\frac{1 + 2}{2 + 3}$ , is a common wrong answer: fractions are added over a common denominator.

Example.

7. When is an even root real? For which values of $x$ is $\sqrt{2x - 6}$ a real number?

A square root is real only when the number under it is 0 or more. So $2x - 6 \ge 0$ . Find where it is exactly 0:

\begin{aligned} 2x - 6 &= 0 \\ x &= 3 \end{aligned}

A larger $x$ makes $2x - 6$ larger, so the root is real for $x \ge {3}$ . Test a value below, such as $x = 1$ : the radicand is negative.

\begin{aligned} t &= 2 \times 1 - 6 \\ &= -4 \end{aligned}

The graph starts at $x = 3$ and has nothing to its left.

Practice

Answer ACT-style questions on radicals and rational exponents. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise radicals and rational exponents with new numbers