Question

\[\sqrt{7} \div 2\sqrt{5} - \sqrt{7}\] rationalize the denominator.

Answer 1

assuming this is
\[\frac{\sqrt{7}}{2\sqrt{5}-\sqrt{7}}\] then multiply top and bottom by the conjugate of the denominator, in other words
\[\frac{\sqrt{7}}{2\sqrt{5}-\sqrt{7}}\times \frac{2\sqrt{5}+\sqrt{7}}{2\sqrt{5}+\sqrt{7}}\]

Answer 2

your denominator will then be \(a^2-b^2=20-7=13\) and your numerator will be whatever you get when you multiply out

Answer 3

I don't get it...

Answer 4

your second post

Answer 5

which step?

Answer 6

oh ok

Answer 7

first of all it is always true that \((a+b)(a-b)=a^2-b^2\)

Answer 8

so what does that make \((2\sqrt{5}-\sqrt{7})(2\sqrt{5}+\sqrt{7})\) ?

Answer 9

it must be
\[(2\sqrt{5})^2-\sqrt{7}^2\]

Answer 10

when you square the first term you get
\[2\sqrt{5}\times 2\sqrt{5}=4\times 5=20\] and when you square the second term you get
\[\sqrt{7}\times \sqrt{7}=7\]

Answer 11

and therefore \((2\sqrt{5}-\sqrt{7})(2\sqrt{5}+\sqrt{7})=(2\sqrt{5})^2-\sqrt{7}^2=20-7=13\)

Answer 12

is that better?

Answer 13

yeaah. thanks!

Answer 14

yw

Answer 15

you still have to multiply out in the numerator don't forget, but that is just the distributive law