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Rationalize 5 over square root of 6?
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.
How do you rationalize 4 with square root if 2?
4=(sqrt2)4
How do you rationalize 5 square root of 3 divided by 2 minus square root of 3?
[5sqrt(3)]/[2 - sqrt(3)] rationalize with the conjugate---- 2 + sqrt(3) ----- A polynomial expansion om bottom and distribution on top [5sqrt(3)]/[2 - sqrt(3)] * 2 + sqrt(3)/2 + sqrt(3) 10sqrt(3) + 5 * 3/2 - 3 - 10sqrt(3) + 15 ============
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).
What is square of square root of 25?
The square root of 25 is 5.
