Math Problem Statement

6. \frac{3\sqrt{5}}{\sqrt{6} - \sqrt{8}} 7. \frac{7}{2\sqrt{3}} + \sqrt{2} 8. \frac{5\sqrt{10}}{\sqrt{7}}

Solution

Let's go through the problems one by one:

Problem 6:

3568\frac{3\sqrt{5}}{\sqrt{6} - \sqrt{8}}

To simplify this, multiply both the numerator and denominator by the conjugate of the denominator, which is 6+8\sqrt{6} + \sqrt{8}. This eliminates the radicals in the denominator:

35686+86+8=35(6+8)(6)2(8)2\frac{3\sqrt{5}}{\sqrt{6} - \sqrt{8}} \cdot \frac{\sqrt{6} + \sqrt{8}}{\sqrt{6} + \sqrt{8}} = \frac{3\sqrt{5}(\sqrt{6} + \sqrt{8})}{(\sqrt{6})^2 - (\sqrt{8})^2}

The denominator simplifies as follows: (6)2(8)2=68=2(\sqrt{6})^2 - (\sqrt{8})^2 = 6 - 8 = -2

Now, distribute in the numerator: 35(6+8)=330+340=330+6103\sqrt{5}(\sqrt{6} + \sqrt{8}) = 3\sqrt{30} + 3\sqrt{40} = 3\sqrt{30} + 6\sqrt{10}

So the expression becomes: 330+6102=33026102=3302310\frac{3\sqrt{30} + 6\sqrt{10}}{-2} = -\frac{3\sqrt{30}}{2} - \frac{6\sqrt{10}}{2} = -\frac{3\sqrt{30}}{2} - 3\sqrt{10}

Problem 7:

723+2\frac{7}{2\sqrt{3}} + \sqrt{2}

To rationalize the first term, multiply the numerator and denominator by 3\sqrt{3}:

72333=7323=736\frac{7}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{2 \cdot 3} = \frac{7\sqrt{3}}{6}

Thus, the expression becomes: 736+2\frac{7\sqrt{3}}{6} + \sqrt{2}

Problem 8:

5107\frac{5\sqrt{10}}{\sqrt{7}}

Rationalize the denominator by multiplying both the numerator and denominator by 7\sqrt{7}:

510777=5707\frac{5\sqrt{10}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{5\sqrt{70}}{7}

Thus, the simplified form is: 5707\frac{5\sqrt{70}}{7}


Would you like further clarification on any of these steps?

Here are 5 related questions:

  1. How do you multiply conjugates involving square roots?
  2. How do you rationalize a denominator with multiple terms?
  3. What happens when you multiply square roots with different radicands?
  4. How can you simplify radicals like 40\sqrt{40} to lowest terms?
  5. Why is rationalizing the denominator important in simplifying expressions?

Tip: Always check if the radicand (the number inside the square root) can be simplified before rationalizing!

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Math Problem Analysis

Mathematical Concepts

Algebra
Radicals
Rationalization of Denominators
Square Roots

Formulas

Conjugate: (a - b)(a + b) = a^2 - b^2
Rationalization: \frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}

Theorems

Conjugate multiplication to rationalize square root denominators
Simplification of radicals

Suitable Grade Level

Grades 9-12