Question - Rationalizing Denominators of Square Roots
Solution:
To solve the given expression, we can use the technique of rationalizing the denominator. The expression given is:$$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}$$We will multiply both the numerator and denominator by the conjugate of the denominator to eliminate the square roots in the denominator. The conjugate of $$\sqrt{2} - \sqrt{3}$$ is $$\sqrt{2} + \sqrt{3}$$. Here's how it is done:$$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \times \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{(\sqrt{2} + \sqrt{3})^2}{(\sqrt{2})^2 - (\sqrt{3})^2}$$Expanding the numerator and simplifying the denominator:$$\frac{2 + 2\sqrt{2}\sqrt{3} + 3}{2 - 3} = \frac{5 + 2\sqrt{6}}{-1}$$Simplifying further:$$\frac{5 + 2\sqrt{6}}{-1} = -5 - 2\sqrt{6}$$The final simplified result is:$$-5 - 2\sqrt{6}$$
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