Math Problem Statement

simplify leaving your answer in the form a+b root c root 5 all over 2root2-root5 + root 2 all over 2root2+root 5

Solution

To simplify the expression you provided:

a+bc225+222+5\frac{a + b \sqrt{c}}{2\sqrt{2} - \sqrt{5}} + \frac{\sqrt{2}}{2\sqrt{2} + \sqrt{5}}

We will start by simplifying both terms. We’ll tackle them one by one, using rationalization for the denominators.

Step 1: Rationalize the first term

The first term is:

a+bc225\frac{a + b \sqrt{c}}{2\sqrt{2} - \sqrt{5}}

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is 22+52\sqrt{2} + \sqrt{5}:

a+bc225×22+522+5=(a+bc)(22+5)(22)2(5)2\frac{a + b \sqrt{c}}{2\sqrt{2} - \sqrt{5}} \times \frac{2\sqrt{2} + \sqrt{5}}{2\sqrt{2} + \sqrt{5}} = \frac{(a + b \sqrt{c})(2\sqrt{2} + \sqrt{5})}{(2\sqrt{2})^2 - (\sqrt{5})^2}

Step 2: Simplify the denominator

Now, let's simplify the denominator:

(22)2=4×2=8(2\sqrt{2})^2 = 4 \times 2 = 8 (5)2=5(\sqrt{5})^2 = 5 So, the denominator becomes:

85=38 - 5 = 3

Step 3: Expand the numerator

Next, expand the numerator:

(a+bc)(22+5)=a(22)+a(5)+bc(22)+bc(5)(a + b \sqrt{c})(2\sqrt{2} + \sqrt{5}) = a(2\sqrt{2}) + a(\sqrt{5}) + b \sqrt{c} (2\sqrt{2}) + b \sqrt{c} (\sqrt{5})

=2a2+a5+2b2c+b5c= 2a\sqrt{2} + a\sqrt{5} + 2b \sqrt{2c} + b \sqrt{5c}

Thus, the first term simplifies to:

2a2+a5+2b2c+b5c3\frac{2a\sqrt{2} + a\sqrt{5} + 2b \sqrt{2c} + b \sqrt{5c}}{3}

Step 4: Rationalize the second term

The second term is:

222+5\frac{\sqrt{2}}{2\sqrt{2} + \sqrt{5}}

We rationalize by multiplying by the conjugate 2252\sqrt{2} - \sqrt{5}:

222+5×225225=2(225)(22)2(5)2\frac{\sqrt{2}}{2\sqrt{2} + \sqrt{5}} \times \frac{2\sqrt{2} - \sqrt{5}}{2\sqrt{2} - \sqrt{5}} = \frac{\sqrt{2}(2\sqrt{2} - \sqrt{5})}{(2\sqrt{2})^2 - (\sqrt{5})^2}

The denominator simplifies as before to:

85=38 - 5 = 3

Now expand the numerator:

2(225)=2(22)2(5)=2×210=410\sqrt{2}(2\sqrt{2} - \sqrt{5}) = \sqrt{2}(2\sqrt{2}) - \sqrt{2}(\sqrt{5}) = 2 \times 2 - \sqrt{10} = 4 - \sqrt{10}

Thus, the second term becomes:

4103\frac{4 - \sqrt{10}}{3}

Step 5: Combine the terms

Now, we can combine both terms, remembering that both denominators are the same:

2a2+a5+2b2c+b5c3+4103\frac{2a\sqrt{2} + a\sqrt{5} + 2b \sqrt{2c} + b \sqrt{5c}}{3} + \frac{4 - \sqrt{10}}{3}

Combine the numerators:

(2a2+a5+2b2c+b5c)+(410)3\frac{(2a\sqrt{2} + a\sqrt{5} + 2b \sqrt{2c} + b \sqrt{5c}) + (4 - \sqrt{10})}{3}

Final Expression:

Thus, the simplified expression is:

2a2+a5+2b2c+b5c+4103\frac{2a\sqrt{2} + a\sqrt{5} + 2b \sqrt{2c} + b \sqrt{5c} + 4 - \sqrt{10}}{3}


Would you like me to provide further details or clarifications on any of these steps? Here are a few related questions you might find interesting:

  1. How do you rationalize a denominator in general?
  2. What happens when we multiply binomials with square roots in them?
  3. How do you combine like terms when dealing with square roots?
  4. What are some strategies for simplifying expressions with mixed square roots and constants?
  5. Can you explain how to handle expressions with square roots in both the numerator and denominator?

Tip: Always check if the denominator is rationalized before simplifying further to avoid complicating the expression!

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Math Problem Analysis

Mathematical Concepts

Rationalizing Denominators
Simplifying Expressions
Algebraic Manipulation
Square Roots

Formulas

Rationalization of square roots
Binomial expansion

Theorems

Conjugate multiplication theorem
Simplification of square roots

Suitable Grade Level

Grades 9-12