imqwerty
  • imqwerty
real math ( ͡~ ͜ʖ ͡°)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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imqwerty
  • imqwerty
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anonymous
  • anonymous
D
imqwerty
  • imqwerty
correct @mukushla :)

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More answers

anonymous
  • anonymous
nice question
anonymous
  • anonymous
we need to find out that \(b^2=ac\)
imqwerty
  • imqwerty
yes correct
ganeshie8
  • ganeshie8
*
Zale101
  • Zale101
Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.
imqwerty
  • imqwerty
well done @Zale
imqwerty
  • imqwerty
i forgot to write 1 step in the solution so heres the new solution
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Sachintha
  • Sachintha
@perl Can you explain it?
perl
  • perl
yes
Sachintha
  • Sachintha
What I didn't get is how \[b^2=ac\]
perl
  • perl
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
Sachintha
  • Sachintha
I still didn't get why \[b^2-ac=0\] Is it taken so that \[\sqrt2\] can be removed?
perl
  • perl
otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational
perl
  • perl
yes to remove the square root 2
perl
  • perl
do you see why b^2 -ac cannot equal to square root 2 ?
perl
  • perl
edit** \[ \Large{ \frac{a\sqrt{2}+b}{b\sqrt{2} + c }\cdot \frac{b\sqrt{2}-c}{b\sqrt{2} -c } \\~\\= \frac{2ab + b^2\sqrt 2 - ac \sqrt 2 - bc}{2b^2 - c^2 } \\~\\= \frac{2ab + \sqrt 2 (b^2 - ac) - bc}{2b^2 - c^2 } }\]
Sachintha
  • Sachintha
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?
perl
  • perl
that part is a bit unclear. we can try to figure it out
Sachintha
  • Sachintha
I think, \[c=ar^2\]
Sachintha
  • Sachintha
r is not defined here :(
Sachintha
  • Sachintha
So b/a is taken as r ?
Sachintha
  • Sachintha
Is it possible to equalize it like that?
Sachintha
  • Sachintha
Well it does work too. :)
Sachintha
  • Sachintha
So we equalize it to another value for the ease of calculations? Sorry for being so dumb.
perl
  • perl
these are good questions :)
perl
  • perl
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
Sachintha
  • Sachintha
I got it. :D Was confused because of r.
perl
  • perl
Yes it looked like it came from out of no where :D
perl
  • perl
In part d) it appears they completed long division.
Sachintha
  • Sachintha
Did long division for myself. :D (whew) It factors to \[(r^2-r+1)(r^2+r+1)\]
perl
  • perl
i see, great job
Sachintha
  • Sachintha
Thank you so much for taking your time and explaining me. :)
perl
  • perl
why can't b^2 -ac = sqrt(2) ?
Sachintha
  • Sachintha
Because we still get a irrational number at the denominator. I am correct?
perl
  • perl
yes can you show why
Sachintha
  • Sachintha
Because \[b^2=\sqrt2+ac\]
perl
  • perl
right
Sachintha
  • Sachintha
Am I wrong?
Sachintha
  • Sachintha
Maybe my answer is not adequate. :D
perl
  • perl
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
perl
  • perl
because of the 2 sqrt(2)
Sachintha
  • Sachintha
yep :)

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