Question

how do you Rationalize the denominator of 4 over quantity of 6 minus square root of 7

Answer 1

@uri Help

Answer 2

Well, you need to multiply both the top and bottom by the conjugate of the denominator. The conjugate is just taking the denominator and changing the sign of it. So this is what you would be solving for:

\[\frac{ 4 }{ (6-\sqrt{7}) }*\frac{ (6+\sqrt{7}) }{{ (6+\sqrt{7}) } }\]

Answer 3

so then id do something like foil?

Answer 4

Right.

Answer 5

okay thankyou :)

Answer 6

Mhm. See what you get and post it if you wanna check your answer :3

Answer 7

alright :)

Answer 8

notice the bottom is in the form (a-b)(a+b) = a^2 + b^2
(it is called the "conjugate" ). the bottom turns into just a nice number

Answer 9

a^2 - b^2

Answer 10

I got a answer but its nothing like what my teacher showed us before

Answer 11

Whatd ya come up with?

Answer 12

4 squareroot 7 +53

Answer 13

Yeah, somehow that didnt turn out. I'll work it out.

Answer 14

\[\frac{ 4 }{ 6-\sqrt{7} }*\frac{ 6+\sqrt{7} }{ 6+\sqrt{7} }\]

Starting with the numerator

\[4(6+\sqrt{7}) = 24 + 4\sqrt{7}\]

Now the denominator I'll work out each step at a time for.

\[6*6 = 36\]
\[6*\sqrt{7} = 6\sqrt{7}\]
\[-\sqrt{7}*6 = -6\sqrt{7}\]
\[-\sqrt{7}*\sqrt{7} = -7\]
From here you think you could fit the rest of the pieces together? :P

Answer 15

let me see what I come up with

Answer 16

okay I got it :) thankyou so much!

Answer 17

Yep, np ^_^

Answer 18

What was mentioned before is a quick way to try and multiply conjugates, though. If you have (a+b)(a-b), you can just find a^2 - b^2 and that would be the answer. So for your problem, I could have just done 6^2 minus sqrt(7)^2 and got the same thing

Answer 19

fix:
notice the bottom is in the form (a-b)(a+b) = a^2 - b^2
(it is called the "conjugate" ). the bottom turns into just a nice number