rationalize the denominator: 3 divide square[6] - square[3]?

Answer 1

3 divide square[6] - square[3]
3 / (\/6 - \/3) =
3(\/6 + \/3) over [(\/6²- \/3²)] =
3(\/6 + \/3) over 6- 3 =
3(\/6 + \/3) over 3 =
\/6 + \/3
The answer is \/6 + \/3.

Answer 2

So you want to rationalize

3 / [ sqrt(6) - sqrt(3) ]

The secret here to rational this is to multiply the denominator by its conjugate. The conjugate of (a - b) is (a + b), and multiplied together, will give a difference of squares, (a^2 - b^2). Note that this is VERY convenient because squaring a radical means the square root sign goes away.

But we can't just multiply the denominator by this; we have to multiply the numerator by this as well, otherwise we'll be changing the expression.

3 [sqrt(6) + sqrt(3)] / {[sqrt(6) - sqrt(3)][sqrt(6) + sqrt(3)]}

Changing the denominator to a difference of squares,

3[sqrt(6) + sqrt(3)] / (6 - 3)
3[sqrt(6) + sqrt(3)] / 3

And now, notice the 3 on the top and bottom will cancel each other out, leaving

sqrt(6) + sqrt(3)

Answer 3

Normally, you want to rationalize the denominator when it is not a rational number. In this case, square(6) - square(3) is a rational number.

I suspect you mean "square root of 6 minus square root of 3"

I use SQRT for square root.

If that is the case, then use these rules:

1) multiply by 1 (which does not change the value). Any non-zero value of the form a/a is equal to 1.
2) "conjugates" are helpful. In this case, if your denominator is SQRT(6) - SQRT(3); the conjugate is SQRT(6) + SQRT(3).

So multiply top and bottom by the conjugate and see what happens.

3*(SQRT(6)+SQRT(3)) / ( (SQRT(6)-SQRT(3))*(SQRT(6)+SQRT(3)) =
3*(SQRT(6)+SQRT(3)) / (6-3) =
3*(SQRT(6)+SQRT(3)) / 3 at this stage, it is rationalized, but we can continue
= SQRT(6)+SQRT(3) (no more denominator!)

= SQRT(2)*SQRT(3) + SQRT(3)
=SQRT(3)*(SQRT(2)+1)

Answer 4

if square [6] - square [3] is the denominator; then
3/((6^(1/2)-3^(1/2)) = 3((6^(1/2)+(3^(1/2))/(6 - 3)
=6^(1/2)+3^(1/2)

if square [6] is the denominator;then
(3/6^(1/2)) -3^(1/2) = (3-18^(1/2))/(6^(1/2))
=(3(6^(1/2))-6(3^(1/2)))/6

Answer 5

3 divided by sqrt6-sqrt3
=(3)(sqrt 6 + sqrt 3)/ (sqrt6-sqrt 3)(sqrt 6+sqrt 3)
=3(sqrt 6 +sqrt 3)/(6-3)
=3(sqrt6+sqrt 3)/3
sqrt 6 +sqrt 3 ans
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Answer 6

i already answered one for you, do you even try to do your homework????