1Cademy - Simplifying $$\frac{\sqrt{x}+\sqrt{7}}{\sqrt{x}-\sqrt{7}}$$

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Rationalizing the Denominator

Example

Simplifying $\frac{\sqrt{x}+\sqrt{7}}{\sqrt{x}-\sqrt{7}}$

Rationalize a two-term denominator where both terms are square roots and the numerator is the conjugate of the denominator.

$\frac{\sqrt{x}+\sqrt{7}}{\sqrt{x}-\sqrt{7}}$

Step 1 — Multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{x}-\sqrt{7}$ is $\sqrt{x}+\sqrt{7}$ :

$\frac{(\sqrt{x}+\sqrt{7})(\sqrt{x}+\sqrt{7})}{(\sqrt{x}-\sqrt{7})(\sqrt{x}+\sqrt{7})}$

Step 2 — Apply the Product of Conjugates Pattern in the denominator. The denominator is a product of conjugates with $a = \sqrt{x}$ and $b = \sqrt{7}$ , so $(\sqrt{x}-\sqrt{7})(\sqrt{x}+\sqrt{7}) = (\sqrt{x})^2 - (\sqrt{7})^2$ :

$\frac{(\sqrt{x}+\sqrt{7})(\sqrt{x}+\sqrt{7})}{(\sqrt{x})^2 - (\sqrt{7})^2}$

Step 3 — Simplify the denominator. Since $(\sqrt{x})^2 = x$ and $(\sqrt{7})^2 = 7$ :

$\frac{(\sqrt{x}+\sqrt{7})^2}{x - 7}$

The numerator is intentionally left in factored form as $(\sqrt{x}+\sqrt{7})^2$ rather than expanded. In this form it is clear that the numerator and denominator share no common factors, so the expression is fully simplified. This example illustrates that when the numerator of the original fraction equals the conjugate used for rationalization, the numerator becomes a perfect square — but expanding it is unnecessary if no cancellation with the denominator is possible.

Updated 2026-04-21

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OpenStax Elementary Algebra

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