real math ( ͡~ ͜ʖ ͡°)

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real math ( ͡~ ͜ʖ ͡°)

Mathematics
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D
correct @mukushla :)

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nice question
we need to find out that \(b^2=ac\)
yes correct
*
Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.
well done @Zale
i forgot to write 1 step in the solution so heres the new solution
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@perl Can you explain it?
yes
What I didn't get is how \[b^2=ac\]
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
I still didn't get why \[b^2-ac=0\] Is it taken so that \[\sqrt2\] can be removed?
otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational
yes to remove the square root 2
do you see why b^2 -ac cannot equal to square root 2 ?
edit** \[ \Large{ \frac{a\sqrt{2}+b}{b\sqrt{2} + c }\cdot \frac{b\sqrt{2}-c}{b\sqrt{2} -c } \\~\\= \frac{2ab + b^2\sqrt 2 - ac \sqrt 2 - bc}{2b^2 - c^2 } \\~\\= \frac{2ab + \sqrt 2 (b^2 - ac) - bc}{2b^2 - c^2 } }\]
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?
that part is a bit unclear. we can try to figure it out
I think, \[c=ar^2\]
r is not defined here :(
So b/a is taken as r ?
Is it possible to equalize it like that?
Well it does work too. :)
So we equalize it to another value for the ease of calculations? Sorry for being so dumb.
these are good questions :)
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
I got it. :D Was confused because of r.
Yes it looked like it came from out of no where :D
In part d) it appears they completed long division.
Did long division for myself. :D (whew) It factors to \[(r^2-r+1)(r^2+r+1)\]
i see, great job
Thank you so much for taking your time and explaining me. :)
why can't b^2 -ac = sqrt(2) ?
Because we still get a irrational number at the denominator. I am correct?
yes can you show why
Because \[b^2=\sqrt2+ac\]
right
Am I wrong?
Maybe my answer is not adequate. :D
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
because of the 2 sqrt(2)
yep :)

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