Question

rationalize the denominator.

\\\frac{\sqrt{3} - 4\sqrt{2}}{7\sqrt{3} + 9\sqrt{2}}\\

\\\frac{\sqrt{3} - 4\sqrt{2}}{7\sqrt{3} + 9\sqrt{2}} = \square\\
(simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

⚡ Using what you learned: Rationalizing Denominators

Step 1: Multiply by the conjugate

$$ \frac{\sqrt{3} - 4\sqrt{2}}{7\sqrt{3} + 9\sqrt{2}} \cdot \frac{7\sqrt{3} - 9\sqrt{2}}{7\sqrt{3} - 9\sqrt{2}} $$

Step 2: Expand the numerator

$$ (\sqrt{3} - 4\sqrt{2})(7\sqrt{3} - 9\sqrt{2}) = 7(3) - 9\sqrt{6} - 28\sqrt{6} + 36(2) $$

$$ = 21 - 37\sqrt{6} + 72 $$

$$ = 93 - 37\sqrt{6} $$

Step 3: Expand the denominator

$$ (7\sqrt{3} + 9\sqrt{2})(7\sqrt{3} - 9\sqrt{2}) = 49(3) - 81(2) $$

$$ = 147 - 162 $$

$$ = -15 $$

Step 4: Simplify the fraction

$$ \frac{93 - 37\sqrt{6}}{-15} = \frac{-93 + 37\sqrt{6}}{15} $$

Answer:

$$ \frac{-93 + 37\sqrt{6}}{15} $$