You cannot. The square root of 5 is irrational.
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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.
4=(sqrt2)4
[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 ============
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).
The square root of 25 is 5.