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How do you rationalize the denominator in this expression 3 over quantity of 6 minus square root of 5?
Wiki User
∙ 13y ago
Multiply the expression (fraction) by (6 + √5)/(6 + √5) (which is 1 and will not change its value).
3/(6 - √5) x (6 + √5)/(6 + √5)
= 3(6 + √5)/(6 - √5)(6 + √5)
= 3(6 + √5)/(36-5) = 3(6 + √5)/31
= 3/31(6 + √5)
Remember:
(a + b)(a - b) = a2 - ab + ab - b2
= a2 - b2
So if either a or b contains a square root, a2 and b2 will not, and so a2 - b2 will not!
Wiki User
∙ 13y ago
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The idea is to get rid of the square root in the denominator. For this purpose, you must multiply numerator and denominator by the square root of 6 in this case.
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Yes. For example, the conjugate of (square root of 2 + square root of 3) is (square root of 2 - square root of 3).
Can you use conjugates to rationalize the denominator even when the denominator contains two radical terms?
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What is the square root of helium?
Square root is a mathematical function whose argument needs to be a mathematical quantity or expression. Since helium is neither a mathematical quantity nor an expression, the question makes no sense.
How do you rationalize denominator surds?
An example may help. If you have the fraction 1 / (2 + root(3)), where root() is the square root function, you multiply top and bottom by (2 - root(3)). If you multiply everything out, you will have no square root in the denominator, instead, you will have a square root in the numerator. If the denominator is only a root, eg root(3), you multiply top and bottom by root(3).
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