The irrational expressions are listed below:
- Case 1: - 5√3 + 6√2
- Case 2: √9 - 1
How to simplify irrational expressions by rationalization
Herein we find two cases of irrational expressions by rationalization, that is, an algebra-based method to eliminate irrational denominators. Both expression can be rationalized by taking advantage of a variation from the following algebra operation:
a² - b² = (a + b) · (a - b)
Now we proceed to simplify each expression:
Case 1
- 3 / (√3 + √2) - 3√2 / (√6 + √3) + 4√3 / (√6 + √2) Given
[- 3 · (√3 - √2)] - [3√2 · (√6 - √3)] / 3 + [4√3 · (√6 - √2)] / 4 Modulative property and associative properties / Existence of multiplicative inverse / Difference of squares
- 3√3 + 3√2 - √12 + √6 + √18 - √6 Multiplication of square roots / Modulative property and associative properties / Existence of multiplicative inverse / Difference of squares
- 3√3 + 3√2 - 2√3 + √6 + 3√2 - √6 Power and root properties
- 5√3 + 6√2 Distributive, associative and modulative properties / Existence of additive inverse / Definitions of addition and subtraction / Result
Case 2
The expression is equivalent to the following sum:
s = ∑ [1 / [√n + √(n + 1)]], for n ∈ {1, 2, 3, 4, ..., 8}, which is equivalent to the following expression:
s = ∑ [√(n + 1) - √n], for n ∈ {1, 2, 3, 4, ..., 8}
Therefore, we find that the result of the finite sum is s = √9 - 1.
To learn more on irrational numbers: https://brainly.com/question/4031928
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