Subjects algebra

Simplify Radical Fractions

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Simplify Radical Fractions


1. **State the problem:** Simplify the expressions \(\frac{7 \sqrt{3} + \sqrt{5}}{\sqrt{5} - 2}\) and \(\frac{9 \sqrt{5} + 3}{4 - \sqrt{10}}\). 2. **Simplify the first expression:** \(\frac{7 \sqrt{3} + \sqrt{5}}{\sqrt{5} - 2}\). Multiply numerator and denominator by the conjugate of the denominator to rationalize it: $$\frac{7 \sqrt{3} + \sqrt{5}}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{(7 \sqrt{3} + \sqrt{5})(\sqrt{5} + 2)}{(\sqrt{5})^2 - (2)^2}$$ Evaluate the denominator: $$(\sqrt{5})^2 - 2^2 = 5 - 4 = 1$$ Expand the numerator: $$7 \sqrt{3} \times \sqrt{5} + 7 \sqrt{3} \times 2 + \sqrt{5} \times \sqrt{5} + \sqrt{5} \times 2 = 7 \sqrt{15} + 14 \sqrt{3} + 5 + 2 \sqrt{5}$$ So the expression becomes: $$7 \sqrt{15} + 14 \sqrt{3} + 5 + 2 \sqrt{5}$$ 3. **Simplify the second expression:** \(\frac{9 \sqrt{5} + 3}{4 - \sqrt{10}}\). Multiply numerator and denominator by the conjugate of the denominator: $$\frac{9 \sqrt{5} + 3}{4 - \sqrt{10}} \times \frac{4 + \sqrt{10}}{4 + \sqrt{10}} = \frac{(9 \sqrt{5} + 3)(4 + \sqrt{10})}{4^2 - (\sqrt{10})^2}$$ Evaluate denominator: $$16 - 10 = 6$$ Expand numerator: $$9 \sqrt{5} \times 4 + 9 \sqrt{5} \times \sqrt{10} + 3 \times 4 + 3 \times \sqrt{10} = 36 \sqrt{5} + 9 \sqrt{50} + 12 + 3 \sqrt{10}$$ Simplify \(\sqrt{50} = 5 \sqrt{2}\): $$36 \sqrt{5} + 9 \times 5 \sqrt{2} + 12 + 3 \sqrt{10} = 36 \sqrt{5} + 45 \sqrt{2} + 12 + 3 \sqrt{10}$$ Divide each term by 6: $$\frac{36 \sqrt{5}}{6} + \frac{45 \sqrt{2}}{6} + \frac{12}{6} + \frac{3 \sqrt{10}}{6} = 6 \sqrt{5} + \frac{15}{2} \sqrt{2} + 2 + \frac{1}{2} \sqrt{10}$$ 4. **Final answers:** First expression: $$7 \sqrt{15} + 14 \sqrt{3} + 5 + 2 \sqrt{5}$$ Second expression: $$6 \sqrt{5} + \frac{15}{2} \sqrt{2} + 2 + \frac{1}{2} \sqrt{10}$$