Radical Functions & Rational Exponents

$\sqrt[n]{a} = b$ exactly when $b^n = a$. Here $n$ is the index and $a$ is the radicand.

• [leftmargin=*,itemsep=1pt]
• Odd index ($\sqrt[3]{\ },\ \sqrt[5]{\ }$): every real number has exactly one real root, and negatives are fine. $\sqrt[3]{-8} = -2$.
• Even index ($\sqrt{\ },\ \sqrt[4]{\ }$): needs a nonnegative radicand, and the principal root is nonnegative. $\sqrt{-9}$ is not real.

To simplify, factor out the largest perfect $n$th power: $\sqrt[n]{a\cdot b}=\sqrt[n]{a}\,\sqrt[n]{b}$.

A fractional exponent is a compact radical. The denominator is the index; the numerator is the power:

Special case $a^{1/n} = \sqrt[n]{a}$. A negative exponent means reciprocal: $a^{-m/n} = \dfrac{1}{a^{m/n}}$.

Every exponent law you know still holds --- just add, subtract, or multiply the fractions:

Give a final answer with positive exponents and one copy of each base.

• [leftmargin=*,itemsep=1pt]
• Add / subtract: only like radicals (same index and same radicand) combine, like adding like terms. Simplify first to reveal them.
• Multiply: $\sqrt[n]{a}\,\sqrt[n]{b}=\sqrt[n]{ab}$; multiply the outside numbers together and the inside numbers together.
• Divide: $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}$, then rationalize so no radical is left in a denominator.

To solve, isolate the radical, then raise both sides to the matching power to undo it.

• [leftmargin=*,itemsep=1pt]
• Get the radical alone on one side.
• Raise both sides to the index (square a square root, cube a cube root).
• Solve the resulting equation.
• Check every answer in the original equation.

Squaring can create extraneous solutions --- values that satisfy the squared equation but not the original. For $x^{m/n}=c$, raise both sides to $n/m$.

• [leftmargin=*,itemsep=1pt]
• $y=\sqrt{x}$: starts at the origin and rises slowly. Domain $[0,\infty)$, range $[0,\infty)$. Passes through $(0,0),(1,1),(4,2),(9,3)$.
• $y=\sqrt[3]{x}$: passes through the origin, defined for all reals, with an S-shape. Domain $(-\infty,\infty)$, range $(-\infty,\infty)$. Passes through $(-8,-2),(-1,-1),(0,0),(1,1),(8,2)$.

Transformations of $y=a\sqrt{x-h}+k$: $h$ shifts right, $k$ shifts up, $a<0$ flips vertically. For a square root, the graph begins at the point $(h,k)$.

Radical functions are the inverses of power functions. Since $\sqrt{x}$ undoes $x^2$ and $\sqrt[3]{x}$ undoes $x^3$:

The restriction $x\ge 0$ on $x^2$ is what makes it one-to-one so an inverse exists. To find an inverse: swap $x$ and $y$, then solve for $y$.

Sometimes a radical hidden inside another radical can be “unpacked” into a sum of two simpler square roots. The goal is to write

Squaring the right side gives $x+y+2\sqrt{xy}$, so matching pieces forces

That is exactly a sum-and-product system: $x$ and $y$ are the two roots of $t^2 - a\,t + \frac{b}{4}=0$. The denesting is clean (nice numbers) only when $a^2-b$ is a perfect square. If it is not, leave the radical nested.

An equation like $x^{2/3}-x^{1/3}-6=0$ looks exotic, but the exponents are in a $2:1$ ratio, so a $u$-substitution turns it into a plain quadratic. Let $u$ be the piece with the smaller exponent:

The equation becomes $u^2-u-6=0$. Solve for $u$, then back-substitute and undo the exponent by raising to the reciprocal power. Because we eventually cube (an odd power) no extraneous roots appear here --- but always confirm the base is allowed (even-index radicals need a nonnegative radicand).

When two radicals appear, you cannot clear both in one squaring. The strategy:

• [leftmargin=*,itemsep=1pt]
• Move one radical to each side (or isolate the messier one).
• Square both sides --- one radical survives, sitting in the cross term.
• Isolate that remaining radical and square a second time.
• Solve, then check every candidate in the original equation.

Two squarings make extraneous solutions especially likely, so the final check is non-negotiable.

Expressions built from $\sqrt{x}$ and its reciprocal are symmetric: swapping $\sqrt{x}\leftrightarrow\frac{1}{\sqrt{x}}$ leaves them unchanged. Let $s=x^{1/2}+x^{-1/2}=\sqrt{x}+\dfrac{1}{\sqrt{x}}$. Squaring links the sum to the middle term:

The constant $2$ appears because the cross term $\sqrt{x}\cdot\frac{1}{\sqrt{x}}=1$. This trick collapses awkward $\sqrt{x}$ equations into quadratics in $s$, and it is the same idea behind $x+x^{-1}$ appearing all over precalculus.

A two-term denominator clears with its conjugate. A three-term denominator such as $\sqrt{a}+\sqrt{b}+\sqrt{c}$ needs grouping: treat two terms as a block, multiply by the conjugate of that block, and repeat.

Now the denominator is a two-term expression $(a+b-c)+2\sqrt{ab}$, which you finish off with one more conjugate. The strategy is always: shrink the number of radical terms by one at each step.

For a radical function, the domain of the function is the range of its inverse, and vice versa --- reflecting across $y=x$ swaps the two axes.

• [leftmargin=*,itemsep=1pt]
• Even-index radical $y=\sqrt{g(x)}$: domain requires $g(x)\ge 0$; the output is nonnegative, so range starts at the minimum output.
• Inverse of a square root is a restricted parabola. The restriction it inherits (e.g. $x\ge k$) comes from the range of the original radical.
• Odd-index radical: domain and range are both all reals; its inverse (a cubic) is unrestricted too.