9.4: Rationalize Algebraic Fractions
If the denominator of a rational expression contains sums or differences involving radicals, it is good form to always rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.
The conjugate of the denominator contains the same terms, but opposite operations (addition or subtraction).
Rationalize the Denominator and Simplify:
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{1}{\sqrt{x} − \sqrt{y}}\)
- \(\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} − \sqrt{y}}\)
Solution
- \(\begin{array} &&\dfrac{1}{1 − \sqrt{x}} &\text{Example problem} \\ &\dfrac{(1)(1 + \sqrt{x})}{(1 − \sqrt{x})(1 + \sqrt{x})} &\text{Multiply both numerator and denominator by the conjugate, which is \((1+\sqrt{x})\)} \\ &\dfrac{1 + \sqrt{x}}{1 − \sqrt{x} + \sqrt{x} − (\sqrt{x})^2} &\text{FOIL the denominator.} \\ &\dfrac{1 + \sqrt{x}}{1 − \cancel{\sqrt{x}} + \cancel{\sqrt{x}} − (\sqrt{x})^2} &\text{Remove opposite terms that sum to zero.} \\ &\dfrac{1 + \sqrt{x}}{1 − x} &\text{The square root of \(x\), quantity squared is \(x\).} \\ &\dfrac{1 + \sqrt{x}}{1 − x} &\text{Final answer with the denominator rationalized, meaning that there are no square root terms in the denominator.} \end{array}\)
- \(\begin{array} &&\dfrac{1}{\sqrt{x} − \sqrt{y}} &\text{Example problem} \\ &\dfrac{(1)(\sqrt{x} + \sqrt{y})}{(\sqrt{x} − \sqrt{y})(\sqrt{x} + \sqrt{y})} &\text{Multiply both numerator and denominator by the conjugate, which is \((\sqrt{x} + \sqrt{y})\)} \\ &\dfrac{(\sqrt{x} + \sqrt{y})}{(\sqrt{x})^2 − \sqrt{x}\sqrt{y} + \sqrt{x}\sqrt{y} − (\sqrt{y})^2} &\text{FOIL the denominator.} \\ &\dfrac{(\sqrt{x} + \sqrt{y}) }{(\sqrt{x})^2 −\cancel{\sqrt{x}\sqrt{y}} + \cancel{\sqrt{x}\sqrt{y}} − (\sqrt{y})^2} &\text{Remove opposite terms that sum to zero.} \\ &\dfrac{\sqrt{x} + \sqrt{y}}{x − y} &\text{The square root of \(x\), quantity squared is \(x\), and the square root of \(y\), quantity squared is \(y\).} \\ &\dfrac{\sqrt{x} + \sqrt{y}}{x − y} &\text{Final answer with the denominator rationalized, meaning that there are no square root terms in the denominator.} \end{array}\)
- \(\begin{array} &&\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} − \sqrt{y}} &\text{Example problem} \\ &\dfrac{(\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y})}{(\sqrt{x} − \sqrt{y})(\sqrt{x} + \sqrt{y})} &\text{Multiply both numerator and denominator by the conjugate, which is \((\sqrt{x} + \sqrt{y})\)} \\ &\dfrac{(\sqrt{x})^2 + 2(\sqrt{x}\sqrt{y}) + (\sqrt{y})^2}{(\sqrt{x})^2 − \sqrt{x}\sqrt{y} + \sqrt{x}\sqrt{y} − (\sqrt{y})^2} &\text{FOIL the numerator and the denominator.} \\ &\dfrac{x + 2\sqrt{x}\sqrt{y} + y}{(\sqrt{x})^2 − \cancel{\sqrt{x}\sqrt{y}} + \cancel{\sqrt{x}\sqrt{y}} − (\sqrt{y})^2} &\text{Remove opposite terms that sum to zero.} \\ &\dfrac{x + 2\sqrt{x}\sqrt{y} + y}{x − y} &\text{The square root of \(x\), quantity squared is \(x\), and the square root of \(y\), quantity squared is \(y\).} \\ &\dfrac{x + 2\sqrt{x}\sqrt{y} + y}{x − y} &\text{Final answer with the denominator rationalized, meaning that there are no square root terms in the denominator.} \end{array}\)
Rationalize the Denominator and Simplify:
- \(\dfrac{x}{1 − \sqrt{x}}\)
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{2 \sqrt{x}}{\sqrt{x} − 1}\)
- \(\dfrac{x − 1}{\sqrt{x} − 1}\)