5.6 Radicals, Rational Exponents, & Approximating Irrational Numbers

Radicals, Rational Exponents, & Approximating Irrational Numbers

Section 5.3 introduced square roots, cube roots, and the laws of exponents separately. This section joins them, which is exactly what the Level A standard requires: "rewrite expressions involving radicals and rational exponents using the properties of exponents."

Radicals Are Exponents

A rational exponent — an exponent that is a fraction — is another way of writing a root.

$x^{1/2} = \sqrt{x} \qquad x^{1/3} = \sqrt[3]{x} \qquad x^{1/n} = \sqrt[n]{x}$

When the numerator is not 1, it becomes a power applied to the root:

$x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$

The denominator is the root; the numerator is the power. A mnemonic that survives test pressure: the root is down below.

Always take the root first. $8^{2/3}$ computed as $\sqrt[3]{64}$ also gives 4, but on non-calculator items rooting first keeps the numbers small: $\sqrt[3]{8} = 2$, then $2^2 = 4$.

The Exponent Laws Still Apply

Nothing about fractional exponents changes the rules. That is the whole reason the notation exists.

Worked example. Simplify $\dfrac{\sqrt{x} \cdot \sqrt[4]{x}}{x^{1/4}}$.

$= \frac{x^{1/2} \cdot x^{1/4}}{x^{1/4}} = x^{1/2 + 1/4 - 1/4} = x^{1/2} = \mathbf{\sqrt{x}}$

Simplifying Radicals

A radical is in simplest form when no perfect-square factor remains under the sign. Pull out the largest perfect square you can find.

$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = \mathbf{6\sqrt{2}}$ $\sqrt{200} = \sqrt{100 \cdot 2} = \mathbf{10\sqrt{2}}$ $\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = \mathbf{3\sqrt[3]{2}}$

Radicals with the same radicand combine like terms: $6\sqrt{2} + 10\sqrt{2} = 16\sqrt{2}$. Radicals with different radicands do not: $\sqrt{2} + \sqrt{3}$ cannot be simplified further, and it is emphatically not $\sqrt{5}$.

Rational vs. Irrational Numbers

Two traps TABE exploits:

1. $\sqrt{\text{perfect square}}$ is rational. $\sqrt{49} = 7$ is an integer, hence rational. Only roots of non-perfect powers are irrational.
2. $\frac{22}{7}$ is rational. It is a ratio, so it is rational by definition, even though it approximates the irrational $\pi$.

Locating an Irrational Number

The tested method is trapping between perfect squares, then narrowing — exactly the procedure the DRC specification describes for truncating a decimal expansion.

Where does $\sqrt{2}$ live?

1. $1^2 = 1$ and $2^2 = 4$, so $1 < \sqrt{2} < 2$.
2. Try tenths: $1.4^2 = 1.96$ and $1.5^2 = 2.25$, so $1.4 < \sqrt{2} < 1.5$.
3. Try hundredths: $1.41^2 = 1.9881$ and $1.42^2 = 2.0164$, so $1.41 < \sqrt{2} < 1.42$.

Each round adds one decimal place. Two rounds is plenty for any TABE item.

Where does $\sqrt{55}$ sit? $7^2 = 49$ and $8^2 = 64$, so it is between 7 and 8, and because 55 is nearer 49 than 64, it is a little below the midpoint: $\sqrt{55} \approx 7.4$.

Ordering a mixed list

Order $\sqrt{10}$, $3.2$, $\pi$, and $\tfrac{16}{5}$ from least to greatest.

So: $\pi < \sqrt{10} < 3.2 = \tfrac{16}{5}$. Converting everything to a decimal with three places is the reliable way to order a mixed list under time pressure.

Estimating expressions containing irrationals

The specification names $\pi^2$ as an example. Replace the irrational with its bounds and square:

$3.1 < \pi < 3.2 \;\Rightarrow\; 9.61 < \pi^2 < 10.24$

So $\pi^2 \approx 9.9$ — close enough to identify the right answer choice without a calculator.