Conjugates and Rationalizing Expressions, foundations for calculus lesson

Plug $x = 0$ into $\frac{\sqrt{x + 4} - 2}{x}$ and you get $\frac{0}{0}$ , which tells you nothing. Yet for small $x$ the values sit very close to $0.25$ . The square root is hiding a factor of $x$ , and the conjugate is the tool that brings it into the open. This exact move comes back in limits and in derivatives of square roots.

Key terms

Radical

Definition For a real number $u \ge 0$ , the square root $\sqrt{u}$ is the unique real number $r \ge 0$ with $r^2 = u$ . In the real numbers $\sqrt{u}$ is undefined when $u < 0$ , so an expression containing $\sqrt{u}$ is defined only where the radicand $u$ is nonnegative. For every real $a$ ,

\sqrt{a^2} = \lvert a \rvert,

and for $u \ge 0$ we have $\left(\sqrt{u}\right)^2 = u$ . An expression written with a root sign, such as $\sqrt{x + 4}$ , is called a radical, and the quantity under the sign is its radicand. This lesson works with square roots, the most common radicals.

In plain words The square root sign always means the nonnegative root. The number under it must not be negative.

Example The expression $\sqrt{x - 3}$ is defined exactly for $x \ge 3$ . Also $\sqrt{(-5)^2} = \sqrt{25} = 5$ , which is $\lvert -5 \rvert$ and not $-5$ .

Difference of squares

Definition For all real numbers $a$ and $b$ ,

(a - b)(a + b) = a^2 - b^2.

The middle terms $ab$ and $-ab$ cancel when the product is expanded.

In plain words A sum times the matching difference leaves only the two squares. When one of the terms is a square root, squaring it removes the root.

Example With $a = \sqrt{7}$ and $b = 2$ ,

\left(\sqrt{7} - 2\right)\left(\sqrt{7} + 2\right) = 7 - 4 = 3.

Conjugate

Definition The conjugate of an expression of the form $A + B$ is $A - B$ , and the conjugate of $A - B$ is $A + B$ . Here one or both of $A$ and $B$ are square roots. For $u \ge 0$ and real $c$ ,

\left(\sqrt{u} - c\right)\left(\sqrt{u} + c\right) = u - c^2.

In plain words Keep the two terms and flip the sign between them. Multiplying an expression by its conjugate turns it into a difference of squares, and the square root disappears.

Example The conjugate of $\sqrt{x + 4} - 2$ is $\sqrt{x + 4} + 2$ . For $x \ge -4$ their product is

(x + 4) - 4 = x.

That is the hidden factor of $x$ from the opening.

Rationalizing

Definition Rationalizing a denominator means rewriting a fraction as an equal fraction whose denominator contains no radicals. Rationalizing a numerator means the same for the numerator. Both are done by multiplying the numerator and the denominator by the same nonzero factor, usually the conjugate of the part that contains the radical, or the radical itself when it stands alone.

In plain words You multiply by a form of $1$ , the conjugate over itself, so the value does not change but the root moves to the other part of the fraction.

Example To rationalize $\frac{4}{3 - \sqrt{5}}$ , multiply above and below by $3 + \sqrt{5}$ , which is not zero.

\begin{aligned} \frac{4}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} &= \frac{4\left(3 + \sqrt{5}\right)}{9 - 5} \\[0.4em] &= 3 + \sqrt{5}. \end{aligned}

When the root stands alone it is enough to multiply by it, as in $\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$ .

Equivalent expression

Definition Two expressions in $x$ are equivalent on a set $S$ when both are defined at every $x$ in $S$ and have the same value there. A rewriting step such as multiplying by $\frac{C}{C}$ produces an expression equivalent to the original on the set where the original is defined and $C \ne 0$ . It does not make the new expression equal to the old one at points where the old one was undefined.

In plain words Simplifying should never change a value. It also cannot create a value where the original had none.

Example Rationalizing the numerator of the opening expression uses the conjugate from above.

\begin{aligned} &\frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} \\[0.4em] &\quad = \frac{x}{x\left(\sqrt{x + 4} + 2\right)} \\[0.4em] &\quad = \frac{1}{\sqrt{x + 4} + 2}. \end{aligned}

At $x = 5$ both forms give $\frac{1}{5}$ . At $x = 0$ the new form gives $\frac{1}{4}$ , but the original is still undefined there. The two are equivalent on $[-4, 0) \cup (0, \infty)$ , so the correct statement is

\begin{aligned} \frac{\sqrt{x + 4} - 2}{x} &= \frac{1}{\sqrt{x + 4} + 2}, \\[0.4em] &x \ge -4, \ x \ne 0. \end{aligned}

Domain restriction

Definition Throughout a rationalization, keep the domain of the original expression. The radicands must be nonnegative, the original denominators must be nonzero, and multiplying above and below by a conjugate $C$ is valid only where $C \ne 0$ . If $C$ is never zero on the original domain, no new exclusion appears. If $C$ is zero somewhere, the new form is equal to the original only away from those points, and the original value there has to be found another way.

In plain words Write the restrictions next to the answer. Check whether the conjugate you multiplied by can be zero.

Example In the opening example the conjugate $\sqrt{x + 4} + 2$ is at least $2$ for every $x \ge -4$ , so it never vanishes and only $x = 0$ is excluded, an exclusion that came from the original denominator. Compare $\frac{1}{\sqrt{x} + 1}$ , which is defined for every $x \ge 0$ and equals $\frac{1}{2}$ at $x = 1$ . Multiplying by the conjugate $\sqrt{x} - 1$ , which is zero at $x = 1$ , gives

\begin{aligned} \frac{1}{\sqrt{x} + 1} &= \frac{\sqrt{x} - 1}{x - 1}, \\[0.4em] &x \ge 0, \ x \ne 1. \end{aligned}

The new form is undefined at $x = 1$ even though the original equals $\frac{1}{2}$ there, so the restriction $x \ne 1$ was created by the conjugate and the point $x = 1$ is lost. For a denominator like $\sqrt{x} + 1$ the original form is already the simpler one, and this rewriting is shown only as a warning.

Check yourself

Rationalize the numerator of $\frac{\sqrt{x + 9} - 3}{x}$ and state for which $x$ the two forms are equal.

Show the answer

Multiplying above and below by $\sqrt{x + 9} + 3$ , which is at least $3$ for $x \ge -9$ , gives

\frac{(x + 9) - 9}{x\left(\sqrt{x + 9} + 3\right)} = \frac{1}{\sqrt{x + 9} + 3}.

The forms are equal for $x \ge -9$ with $x \ne 0$ . At $x = 0$ the original is undefined, even though the new form gives $\frac{1}{6}$ .

Rationalize the denominator of $\frac{10}{\sqrt{7} + \sqrt{2}}$ .

Show the answer

Multiply above and below by $\sqrt{7} - \sqrt{2}$ , which is not zero.

\frac{10\left(\sqrt{7} - \sqrt{2}\right)}{7 - 2} = 2\left(\sqrt{7} - \sqrt{2}\right).

Simplify $\frac{\sqrt{x} - 2}{x - 4}$ by using a conjugate, and give the restrictions.

Show the answer

The original needs $x \ge 0$ and $x \ne 4$ . Multiplying above and below by $\sqrt{x} + 2$ , which is never zero, turns the numerator into $x - 4$ , so

\begin{aligned} \frac{\sqrt{x} - 2}{x - 4} &= \frac{1}{\sqrt{x} + 2}, \\[0.4em] &x \ge 0, \ x \ne 4. \end{aligned}

You can also see it directly, since $x - 4 = \left(\sqrt{x} - 2\right)\left(\sqrt{x} + 2\right)$ for $x \ge 0$ .

Conjugates and Rationalizing Expressions

Key terms

Radical

Difference of squares

Conjugate

Rationalizing

Equivalent expression

Domain restriction

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus