Rationalization questions for of having denominators of 3 terms
Answers
Your problem has three terms in the denominator: a + b + c
However, imagine for a moment how you would rationalize a denominator with only two terms: a + b.
As you know, if the denominator contains only two terms, you could rationalize the denominator by multiplying the denominator by its conjugate: a - b.
The difference of squares formula states that:
(a + b)(a − b) = a^2 − b^2
You can apply the same reasoning to rationalize a denominator which contains three terms. GROUP THE TERMS as follows:
a + b + c = (a + b) + c
The difference of squares formula states that:
[(a + b) + c] * [(a + b) - c] = (a + b) ^2 − c^2
Original Problem:
1/(1+3^1/2-5^1/2)
Group the denominator so that the difference of squares formula can be applied:
1/[(1 + 3^1/2) - 5^1/2]
Multiply by the conjugate: [(1 + 3^1/2) + 5^1/2]
1/[(1 + 3^1/2) - 5^1/2]
{1/[(1 + 3^1/2) - 5^1/2]} * {[(1 + 3^1/2) + 5^1/2] / [(1 + 3^1/2) + 5^1/2]}
(1 + 3^1/2 + 5^1/2) / [(1 + 3^1/2)^2 + (5^1/2)^2]
(1 + 3^1/2 + 5^1/2) / (1 + 2*3^1/2 +3 ? 5)
(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)
Repeat the process - Multiply by the new conjugate: (2*3^1/2 + 1)
(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)
[(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)] * [(2*3^1/2 + 1)/ (2*3^1/2 + 1)]
[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / [(2*3^1/2 ? 1) * (2*3^1/2 + 1)]
[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / (2*3^1/2)^2 ? 1^2)
[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / (12 - 1)
[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / 11
(2 * 3^1/2 + 6 + 2 * 15^1/2 + 1 + 3^1/2 + 5^1/2) / 11
(7 + 3 * 3^1/2 + 5^1/2 + 2 * 15^1/2) / 11
The final answer is: (7 + 3 * 3^1/2 + 5^1/2 + 2 * 15^1/2) / 11