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Dividing 1010 by 1.41421...1.41421... by hand would mean computing long division with an endless, non-repeating decimal divisor. But if you rewrite 102\frac{10}{\sqrt{2}} as 525\sqrt{2}, you only have to multiply 55 by 1.41421...1.41421... in seconds.
Rewriting an expression to remove radicals from the bottom of a fraction is called rationalizing the denominator, turning an unruly divisor into a clean rational number.
Multiplying a single square root by itself is straightforward — but how do you eliminate radicals when two different square roots are added together in the denominator?

The Binomial Conjugate

To clear two square roots simultaneously, we exploit the difference of two squares identity: (u−v)(u+v)=u2−v2(u - v)(u + v) = u^2 - v^2. The expression formed by swapping the middle sign is called the algebraic conjugate.
Because squaring undoes a square root, multiplying (a−b)(\sqrt{a} - \sqrt{b}) by (a+b)(\sqrt{a} + \sqrt{b}) squares both terms individually and cancels the radical cross terms entirely.
Let's rationalize 67−3\frac{6}{\sqrt{7} - \sqrt{3}} by multiplying the top and bottom by its conjugate, (7+3)(\sqrt{7} + \sqrt{3}).
The denominator collapses to (7)2−(3)2=7−3=4(\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3 = 4. Multiplying the numerator gives 6(7+3)6(\sqrt{7} + \sqrt{3}), which simplifies to 3(7+3)2\frac{3(\sqrt{7} + \sqrt{3})}{2}.
A common mistake is changing signs inside the radicals rather than between them. The conjugate of 7−3\sqrt{7} - \sqrt{3} is strictly 7+3\sqrt{7} + \sqrt{3}, never −7−−3\sqrt{-7} - \sqrt{-3}.
Difference of squares works miracles for index-2 roots. But what happens if the denominator contains cube roots, where squaring leaves radicals behind?

Higher-Order Binomial Surds

To clear cube roots, terms must be cubed rather than squared, which means applying the sum and difference of cubes identities: (u±v)(u2∓uv+v2)=u3±v3(u \pm v)(u^2 \mp uv + v^2) = u^3 \pm v^3.