Question

Why doesn't anyone help me? ;( is it because of my name How do you rationalize 3 over 3+ radical 7 - 10 over radical 7!! help help help :)

Answer 1

i know an aisha and she is very nice

Answer 2

I am also very nice :)

Answer 3

I feel like no one helps me because of my name. I can change it if anyone likes?

Answer 4

\[\frac{3}{3+\sqrt{7}}=\frac{3}{3+\sqrt{7}}\times\frac{3-\sqrt7}{3-\sqrt7}\]
\[=\frac{3(3-\sqrt7)}{9-7}=\frac{9-3\sqrt{7}}{2}\]

Answer 5

Oh you are the best!!!!

Answer 6

I really appreciate it :) If I ACE that final I owe you one lol

Answer 7

you rationalize the denominator by multiplying top and bottom by the conjugate of the denominator
the conjugate of \(a+\sqrt{b}\) is \(a-\sqrt{b}\) and this works because
\[(a+\sqrt b)(a-\sqrt b)=a^2-b\] so there is no more radical in the denominator

Answer 8

oh okay I got that part but how do you rationalize \[\sqrt{7}\]

Answer 9

for
\[\frac{10}{\sqrt{7}}\] it is easier, just multiply top and bottom by \(\sqrt{7}\) and get
\[\frac{10\sqrt{7}}{7}\]

Answer 10

now your problem is
\[\frac{9-3\sqrt{7}}{2}-\frac{10\sqrt{7}}{7}\] so you need a common denominator, namely \(14\)

Answer 11

OH awesome! so you are left with 9-3\[\sqrt{7}\]/2 - 10\[\sqrt{7}\]/7

Answer 12

\[\frac{7(9-3\sqrt{7})-2(10\sqrt{7})}{14}\] is the first step

Answer 13

then multiply out in the numerator and combine like terms

Answer 14

Oh okay.. wow you're great!

Answer 15

\[\frac{63-21\sqrt{7}-20\sqrt{7}}{14}=\frac{63-41\sqrt{7}}{14}\]

Answer 16

hope all steps are clear

Answer 17

Yes, indeed they are!

Answer 18

ya but I guess that's not the case anymore after satelite helped me