Question

real math ( ͡~ ͜ʖ ͡°)

Answer 1

D

Answer 2

correct @mukushla :)

Answer 3

nice question

Answer 4

we need to find out that \(b^2=ac\)

Answer 5

yes correct

Answer 6

*

Answer 7

Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.

Answer 8

well done @Zale

Answer 9

i forgot to write 1 step in the solution so heres the new solution

Answer 10

@perl Can you explain it?

Answer 11

yes

Answer 12

What I didn't get is how \[b^2=ac\]

Answer 13

We 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

Answer 14

I still didn't get why \[b^2-ac=0\] Is it taken so that \[\sqrt2\] can be removed?

Answer 15

otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational

Answer 16

yes to remove the square root 2

Answer 17

do you see why b^2 -ac cannot equal to square root 2 ?

Answer 18

edit**
\[ \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 }

}\]

Answer 19

Ohh I forgot all about sets of real numbers including rational and irrational .. :D
But what does number are a,ar and ar^2 mean?

Answer 20

that part is a bit unclear. we can try to figure it out

Answer 21

I think, \[c=ar^2\]

Answer 22

r is not defined here :(

Answer 23

So b/a is taken as r ?

Answer 24

Is it possible to equalize it like that?

Answer 25

Well it does work too. :)

Answer 26

So we equalize it to another value for the ease of calculations? Sorry for being so dumb.

Answer 27

these are good questions :)

Answer 28

since 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

Answer 29

I got it. :D Was confused because of r.

Answer 30

Yes it looked like it came from out of no where :D

Answer 31

In part d) it appears they completed long division.

Answer 32

Did long division for myself. :D (whew) It factors to \[(r^2-r+1)(r^2+r+1)\]

Answer 33

i see, great job

Answer 34

Thank you so much for taking your time and explaining me. :)

Answer 35

why can't b^2 -ac = sqrt(2) ?

Answer 36

Because we still get a irrational number at the denominator. I am correct?

Answer 37

yes can you show why

Answer 38

Because \[b^2=\sqrt2+ac\]

Answer 39

right

Answer 40

Am I wrong?

Answer 41

Maybe my answer is not adequate. :D

Answer 42

if we plug in the denominator we get
2b^2 -c^2 = 2 ( √2 + ac) - c^2 = 2√2 + 2ac - c^2 , which is irrational

Answer 43

because of the 2 sqrt(2)

Answer 44

yep :)