The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 18 (Intermediate Algebra) Solving Quadratic Equations with Radicals and Exponents

Overview

This lecture opens the chapter on quadratics by solving equations that involve radicals and rational/integer exponents. The central skill is undoing an operation with its inverse: clear a radical by raising both sides to the matching power, and clear an exponent by taking the matching root. Along the way it draws a sharp distinction between the value of an even root (always positive) and the solutions of an even-power equation (two, written with $\pm$ ). Because squaring (or any even power) can introduce solutions that do not satisfy the original equation, every problem with an even radical or even root must be checked. Odd roots and odd powers behave more simply and need no $\pm$ .

Key Concepts

Even vs. odd radicals (the value of a root). By definition, an even root returns only the non-negative value: $\sqrt{4}=2$ (not $-2$ ), and $\sqrt[4]{16}=2$ . You can never get a negative number out of an even radical. An odd root keeps the sign of its input, so it can be negative: $\sqrt[3]{-8}=-2$ .
A root vs. an equation — where $\pm$ comes from. The symbol $\sqrt{4}$ names a single number, $2$ . But the equation $x^2=4$ has two solutions, $x=\pm 2$ , because both $2$ and $-2$ square to $4$ . The $\pm$ belongs to solving an even-power equation, not to the radical symbol itself.
Solving a radical equation (radical equals a number/expression). Isolate the radical on one side, then raise both sides to the index's power to cancel it: $\big(\sqrt[n]{A}\big)^{\,n}=A$ . Squaring undoes a square root, cubing undoes a cube root, and so on.
Even index, negative target = no solution. If an isolated even radical equals a negative number (e.g. $\sqrt[4]{3x+6}=-3$ ), there is no solution, since an even root is never negative. Algebraically "solving" anyway produces a value that fails the check — a deliberate trap.
Always check even-radical answers. Squaring both sides can create extraneous solutions. After solving, substitute each candidate into the original equation; keep only those that make it true (it is not enough that the radicand stays non-negative — the whole equation must hold). Odd-index equations need no such check.
Two radicals → isolate and square twice. With two radicals, put the harder one alone on one side, square, isolate the remaining radical, then square again. Each squaring removes one radical.
Solving by canceling an exponent (root of both sides). To solve $x^n=k$ , take the $n$ -th root of both sides.
- Odd exponent: one solution, $x=\sqrt[n]{k}$ (no $\pm$ ).
- Even exponent: two solutions, $x=\pm\sqrt[n]{k}$ — anytime you undo an even power you must write $\pm$ (or get no real solution if $k<0$ ).
Rational exponents. A power $a^{m/n}$ means " $n$ -th root, then $m$ -th power" (order is interchangeable). To clear it, raise both sides to the reciprocal exponent $n/m$ , since $\big(a^{m/n}\big)^{n/m}=a^{1}$ . Equivalently, undo it one piece at a time — root then power — applying $\pm$ whenever an even root is undone. Choosing the convenient order (e.g. root first when the radicand is a perfect power) keeps the arithmetic clean.

Worked Example

Solve a two-radical equation, then check (squaring can create extraneous roots): $\sqrt{4x+1}=5-x.$

\begin{aligned} \big(\sqrt{4x+1}\big)^2 &= (5-x)^2 \\ 4x+1 &= 25-10x+x^2 \\ 0 &= x^2-14x+24 \\ 0 &= (x-12)(x-2) \\ x &= 12 \quad\text{or}\quad x = 2. \end{aligned}

Check each candidate in the original equation:

\begin{aligned} x=12:&\quad \sqrt{4(12)+1}=\sqrt{49}=7,\quad 5-12=-7 \;\Rightarrow\; 7\neq -7 \;\text{(reject)} \\ x=2:&\quad \sqrt{4(2)+1}=\sqrt{9}=3,\quad 5-2=3 \;\Rightarrow\; 3=3 \;\text{(keep)}. \end{aligned}

So the only solution is $x=2$ ; $x=12$ is extraneous — exactly why even-radical answers must be checked.

Summary

An even root always returns a non-negative value; an odd root keeps its sign. The $\pm$ comes from solving an even-power equation, not from the radical symbol.
To solve a radical equation, isolate the radical and raise both sides to its index's power. With two radicals, isolate and square twice.
An isolated even radical equal to a negative number has no solution.
After solving any even-radical equation, substitute back into the original equation and discard extraneous solutions; odd-index equations need no check.
To solve $x^n=k$ : odd $n$ gives one root $\sqrt[n]{k}$ ; even $n$ gives $\pm\sqrt[n]{k}$ (or no real solution when $k<0$ ).
Clear a rational exponent $m/n$ by raising both sides to $n/m$ , attaching $\pm$ whenever an even root is undone.