Rationalize
V stands for square root
3/V2
3/V2 x V2/V2 = 3V2/V2V2= 3V2/V4 =3V2/2
Answer 3V2/2
6V2/V3
6V2/3 x V3/V3 =6V2V3/V3V3 = 6V6/3= 2V6
My book gives these two examples. In the first problem you multiply the 2 in the denominator and get 4. It the second problem the 3 in the denominator is not multipled. I am not clear why the 3 is not multiplied in the denominator in the second problem?
In the second, you have 6sqrt(2/3)=sqrt2/sqrt3
multiply numerator and denominator by sqrt3 to get 6(sqrt6)/3= 2sqrt6
I am still not clear?
In the first problem the denominator of 2 was multipled to get the square root of 4.
Why would multiply the 3 in the second problem to get the square root of 9?
In the first problem, the 2 in the denominator (√2) is multiplied by √2 to rationalize the denominator because having a radical (√) in the denominator is not desirable. Multiplying by √2 rationalizes it because it cancels out the radical in the denominator and gives a rational number (√2 x √2 = 2).
However, in the second problem, the 3 in the denominator (√3) is not multiplied by anything to rationalize the denominator. This is because multiplying by √3 would not result in an integer or rational number, as √3 is an irrational number (it cannot be expressed as a simple fraction). So, in this case, the denominator is already in the simplest form possible.
Rationalizing the denominator is done to simplify the expression and remove any radicals from the denominator, but it should be noted that in some cases, like the second example you provided, the denominator may already be in the simplest form without any further simplification.