A community for students.
Here's the question you clicked on:
 0 viewing
imqwerty
 one year ago
real math ( ͡~ ͜ʖ ͡°)
imqwerty
 one year ago
real math ( ͡~ ͜ʖ ͡°)

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we need to find out that \(b^2=ac\)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.1Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2i forgot to write 1 step in the solution so heres the new solution

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1@perl Can you explain it?

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1What I didn't get is how \[b^2=ac\]

perl
 one year ago
Best ResponseYou've already chosen the best response.4We want to force the entire expression to be rational, given a,b,c are integers. You have two possibilities, b^2 ac = 0 or sqrt(2) b^2 ac cannot equal to sqrt(2) , because of the denominator. So you are left with only b^2 ac = 0

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1I still didn't get why \[b^2ac=0\] Is it taken so that \[\sqrt2\] can be removed?

perl
 one year ago
Best ResponseYou've already chosen the best response.4otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational

perl
 one year ago
Best ResponseYou've already chosen the best response.4yes to remove the square root 2

perl
 one year ago
Best ResponseYou've already chosen the best response.4do you see why b^2 ac cannot equal to square root 2 ?

perl
 one year ago
Best ResponseYou've already chosen the best response.4edit** \[ \Large{ \frac{a\sqrt{2}+b}{b\sqrt{2} + c }\cdot \frac{b\sqrt{2}c}{b\sqrt{2} c } \\~\\= \frac{2ab + b^2\sqrt 2  ac \sqrt 2  bc}{2b^2  c^2 } \\~\\= \frac{2ab + \sqrt 2 (b^2  ac)  bc}{2b^2  c^2 } }\]

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Ohh I forgot all about sets of real numbers including rational and irrational .. :D But what does number are a,ar and ar^2 mean?

perl
 one year ago
Best ResponseYou've already chosen the best response.4that part is a bit unclear. we can try to figure it out

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1r is not defined here :(

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1So b/a is taken as r ?

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Is it possible to equalize it like that?

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Well it does work too. :)

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1So we equalize it to another value for the ease of calculations? Sorry for being so dumb.

perl
 one year ago
Best ResponseYou've already chosen the best response.4these are good questions :)

perl
 one year ago
Best ResponseYou've already chosen the best response.4since b^2 = ac , divide both sides by a $$ \Large c = \frac{b^2}{ a} = a \cdot \frac{b^2}{ a^2 } = a \cdot \left(\frac b a \right)^2 $$ Let r = (b/a) Now we have a= a ar = a (b/a) = b ar^2= a (b/a)^2 = c

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1I got it. :D Was confused because of r.

perl
 one year ago
Best ResponseYou've already chosen the best response.4Yes it looked like it came from out of no where :D

perl
 one year ago
Best ResponseYou've already chosen the best response.4In part d) it appears they completed long division.

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Did long division for myself. :D (whew) It factors to \[(r^2r+1)(r^2+r+1)\]

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Thank you so much for taking your time and explaining me. :)

perl
 one year ago
Best ResponseYou've already chosen the best response.4why can't b^2 ac = sqrt(2) ?

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Because we still get a irrational number at the denominator. I am correct?

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Because \[b^2=\sqrt2+ac\]

Sachintha
 one year ago
Best ResponseYou've already chosen the best response.1Maybe my answer is not adequate. :D

perl
 one year ago
Best ResponseYou've already chosen the best response.4if we plug in the denominator we get 2b^2 c^2 = 2 ( √2 + ac)  c^2 = 2√2 + 2ac  c^2 , which is irrational
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.