Question

Square Root Dividing, Please Help!

√81/√7 square root of 81 over square root of 7


2/6-√3 2 over 6 - the square root of 3

Answer 1

\[\frac{\sqrt{81}}{\sqrt7}=\frac{\sqrt{81}}{\sqrt7}\times 1\]\[=\frac{\sqrt{81}}{\sqrt7}\times \frac{\sqrt 7}{\sqrt{7}}\]\[=\frac{{\sqrt{81}\times \sqrt 7}}{7}\]\[=\frac{9\times \sqrt7}{7}\]

Answer 2

The best thing to do here is to rationalise the denominator.

For the first one, √81/√7, if we multiply top and bottom by √7, we eliminate the square root signs.

No hang on a minute, √81 is 9, so √81/√7 = 9/√7 = 9√7 / √7 √7 = 9√7/7.

For the second one, 2/6-√3, multiply top and bottom by the conjugate of the bottom, i.e. reverse the sign.

So 2/6-√3 = (2/6-√3) * (6+√3/6+√3) = 2(6+√3) /(6+√3)(6-√3)

Notice the bottom is the difference of two squares, which is nice :)

So this simplifies to (12 + 2√3)/(36-3) = (12 + 2√3)/33 .

Hope this helps!

Answer 3

Thanks Guys, those are actually what I got for answers.. but I doubt myself a lot and get worked checked.. Thanks!