Multiplying by Conjugates

Core Definition

Multiplying by the conjugate is an algebraic simplification technique used to eliminate radicals (square roots) from the numerator or denominator of a rational expression. It exploits the difference of squares identity:

(A - B)(A + B) = A^2 - B^2

If you have a two-term expression containing a radical, such as

\sqrt{f(x)} - g(x)

, its conjugate is

\sqrt{f(x)} + g(x)

(and vice versa).

Concept Explanation

When we multiply a radical binomial by its conjugate, the square root terms are squared, and the middle cross-product terms cancel out:

(\sqrt{u} - v)(\sqrt{u} + v) = (\sqrt{u})^2 + v\sqrt{u} - v\sqrt{u} - v^2 = u - v^2

Notice that the resulting expression (

u - v^2

) contains no radicals.

To preserve the value of a fraction, we must multiply both the numerator and the denominator by the conjugate expression. This is equivalent to multiplying the fraction by a "sneaky one" (an expression divided by itself, $\frac{D}{D} = 1$ ).

Why We Use It in Calculus

The primary use of conjugate multiplication in Calculus is evaluating limits of rational functions that involve radicals and result in the indeterminate form $\frac{0}{0}$ .

Resolving $\frac{0}{0}$ Limits (Calculus I)

Consider the limit:

\lim_{x \to c} \frac{\sqrt{x + k} - L}{x - c}

Direct substitution yields

\frac{0}{0}

. The radical prevents us from factoring or dividing.

By multiplying the numerator and denominator by the conjugate $\sqrt{x+k} + L$ :

The numerator simplifies to $(x + k) - L^2$ .
Through algebraic setup, $(x+k) - L^2$ will simplify to contain a factor of $(x-c)$ .
This common factor $(x-c)$ cancels out with the $(x-c)$ in the denominator, removing the division-by-zero term.
You can then evaluate the limit by direct substitution.

How to Recognize When to Use It

The Radical Clue: You are trying to simplify a fraction or evaluate a limit, and it contains a binomial expression with a square root (e.g. $\sqrt{x+5} - 3$ or $x - \sqrt{x^2+2}$ ).
The Indeterminate Clue: Plugging the limit value in results in $\frac{0}{0}$ .
No Simple Factoring: The presence of the root prevents standard trinomial factoring or division.

Example Problems

Example Problem 1: Basic Indeterminate Limit

Evaluate the limit:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}

Solution

Step 1: Check for indeterminate form Substitute $x = 4$ :

\frac{\sqrt{4} - 2}{4 - 4} = \frac{2 - 2}{0} = \frac{0}{0}

This is a

\frac{0}{0}

indeterminate form. The radical in the numerator prevents factoring.

Step 2: Find the conjugate The binomial with the radical is $\sqrt{x} - 2$ . Its conjugate is $\sqrt{x} + 2$ .

Step 3: Multiply numerator and denominator by the conjugate

\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}

Expand the numerator using $(A-B)(A+B) = A^2 - B^2$ :

\frac{(\sqrt{x})^2 - 2^2}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}

Note: Keep the denominator factored! Do not distribute the $(x-4)$ through the conjugate, as we want to cancel it.

Step 4: Cancel the common factor and evaluate Cancel $(x-4)$ from both the top and bottom:

\lim_{x \to 4} \frac{1}{\sqrt{x} + 2}

Now evaluate by direct substitution:

\frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}

The limit is $\frac{1}{4}$ .

Example Problem 2: Binomial Root under the Radical

Evaluate the limit:

\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}

Solution

Step 1: Check for indeterminate form Substitute $x = 0$ :

\frac{\sqrt{0 + 9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}

This is a

\frac{0}{0}

indeterminate form.

Step 2: Find the conjugate The radical binomial is $\sqrt{x+9} - 3$ . Its conjugate is $\sqrt{x+9} + 3$ .

Step 3: Multiply numerator and denominator by the conjugate

\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}

Expand the numerator:

\frac{(\sqrt{x + 9})^2 - 3^2}{x(\sqrt{x + 9} + 3)} = \frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \frac{x}{x(\sqrt{x + 9} + 3)}

Step 4: Cancel the common factor and evaluate Cancel $x$ from the numerator and denominator:

\lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3}

Substitute $x = 0$ :

\frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}

The limit is $\frac{1}{6}$ .

Connections & References

Parent Concepts: Calculus Foundations
Related Concepts: Limits, Factoring Second Order Polynomials
Course Links: Calc-1 Discussion 01