Radicals and Rational Exponents

Radicals as rational exponents

This one fact turns every radical rule into an exponent rule you already know: $a^p \cdot a^q = a^{p+q}$, $(a^p)^q = a^{pq}$, $\dfrac{a^p}{a^q} = a^{p-q}$.

Throughout this page, variables represent positive real numbers. That way we never worry about even roots of negative numbers. (Without that assumption, $\sqrt[4]{a^2}$ would be $\sqrt{|a|}$.)

Simplifying radicals

A radical is in simplest form when no factor of the radicand is a perfect $n$th power and the index is as small as possible. Some problems say take out factors from the radical; that is exactly what simplifying means.

A shortcut for variables: divide the exponent by the index. The quotient comes out, the remainder stays in. For $\sqrt{x^7}$, $7 \div 2 = 3$ remainder $1$, so $\sqrt{x^7} = x^3\sqrt{x}$.

Power of a radical. $\left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$. Using exponents, it is easiest to multiply the exponent before simplifying.

Root of a root. $\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$ because $(a^{1/m})^{1/n} = a^{1/(mn)}$. The indices multiply.

Reducing to a common index

How do you compare $\sqrt{3}$ and $\sqrt[3]{5}$? They have different indices, so you cannot compare the radicands directly. Write both with the same index, using the LCM of the indices.

The same trick lets you multiply and divide radicals of different indices: convert to exponents and add or subtract them.

Adding, subtracting, and multiplying radicals

Multiplying radicals with the same index: $\sqrt[n]{a}\cdot\sqrt[n]{b} = \sqrt[n]{ab}$. Multiply the radicands, then simplify.

Special products work with radicals too. Treat each radical like a variable. Remember $(\sqrt{k})^2 = k$.

Rationalizing denominators

By convention, a simplified answer has no radical in the denominator. Rationalizing is what makes $\dfrac{1}{\sqrt{2}}$ become $\dfrac{\sqrt{2}}{2}$. Both are the same number, but the second is easier to add, compare, and estimate by hand.

Three terms. Group two of them so the denominator looks like $A - B$, then use the conjugate $A + B$. You may need to rationalize a second time.

Radicals in geometry and physics

Radicals show up whenever a quantity is squared: the Pythagorean Theorem, the period of a pendulum, and energy and scaling laws. In these problems the answer should be exact and simplified, not a rounded decimal.

Scaling with roots. The period of a pendulum is $T = 2\pi\sqrt{\dfrac{L}{g}}$. Notice $T$ is proportional to $\sqrt{L}$. If you want the period to be $k$ times as large, you need $\sqrt{L'} = k\sqrt{L}$, so square both sides: $L' = k^2 L$.

Fractional exponents in science laws. Many scaling laws look like $R = kM^{3/4}$. If the mass is multiplied by $c$, the quantity is multiplied by $c^{3/4}$.

Back to the start: Emma's box has edges $2$, $3$, and $5$ ft, so the longest rod is $D = \sqrt{4 + 9 + 25} = \sqrt{38}$ ft. Since $38 = 2 \cdot 19$ has no perfect-square factor, $\sqrt{38}$ ft is already in simplest form. That is about $6.16$ ft.

Geometry challenges: heights and spirals

Two classic figures need the Pythagorean Theorem again and again, and the answers are nested or growing radicals.

Heights of an isosceles triangle and trapezoid

If an isosceles triangle has legs $\ell$ and base $b$, the altitude to the base splits the base in half, so $h = \sqrt{\ell^2 - \left(\tfrac{b}{2}\right)^2}$.

An isosceles trapezoid with bases $B > b$ and legs $\ell$ has two right triangles at its ends. Each has horizontal leg $\dfrac{B-b}{2}$ and hypotenuse $\ell$, so $h = \sqrt{\ell^2 - \left(\tfrac{B-b}{2}\right)^2}$.

A regular pentagon with side $L$ has all five diagonals equal to $D = L\Phi$, where $\Phi = \dfrac{1+\sqrt{5}}{2}$ is the golden ratio. A horizontal diagonal cuts it into an isosceles triangle on top (legs $L$, base $D$) and an isosceles trapezoid below (bases $D$ and $L$, legs $L$). The pentagon's height is the sum of the two heights.

The squaring trick is the same one you will meet in the challenge section: to simplify a sum of two radicals, square it and let the cross term $2h_1h_2$ collapse.

The spiral of Theodorus

Start with a right triangle whose two legs are both $a$. Each new right triangle uses the previous hypotenuse as one leg and a new leg of length $a$. The area of a right triangle is $\tfrac{1}{2}\cdot\text{leg}\cdot\text{leg}$.

Let $h_n$ be the hypotenuse of the $n$th triangle. Then $h_n^2 = h_{n-1}^2 + a^2$, and starting from $h_1^2 = 2a^2$ each step adds $a^2$:

The $n$th triangle has legs $h_{n-1} = a\sqrt{n}$ and $a$. Its area and perimeter are

Challenge: nested radicals and clever conjugates

Three techniques cover the hardest problems.

1. Nested radicals with variables. Convert everything to exponents, work from the inside out, and keep a running exponent for each variable.

2. Squaring to unmask a radical sum. To show that something like $\sqrt{A} \pm \sqrt{B}$ is a nice number, set it equal to $x$, square it, and use $(p \pm q)^2 = p^2 \pm 2pq + q^2$. The cross term $2\sqrt{AB}$ often simplifies because $AB$ is a difference of squares.

3. Fractions with conjugate radicals. Combine over a common denominator. The denominator becomes a difference of squares and the radicals vanish.

Summary