rvc
  • rvc
Please Help. I need help with the first two questions. And is the third answer's solution's step is correct?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rvc
  • rvc
hartnn
  • hartnn
know the a^3- b^3 formula?
rvc
  • rvc
yep

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hartnn
  • hartnn
use the property that, in a quadratic equation ax^2+bx+c=0 sum of roots = -b/a product of roots = c/a find the difference of roots from the above two values
hartnn
  • hartnn
difference^2 = sum^2 -4*product
anonymous
  • anonymous
The third one looks correct. So the solutions would be the appropriate roots of unity. For the second one , write \(x_n\) as \(e^{i\frac{\pi}{2^n}}\) and proceed.
rvc
  • rvc
hmmm.. i gtg now too busy atm ill check these sums later. Thanks to all. If i require help i will surely tag u ppl. :) thanks once again
rvc
  • rvc
for the second one?
rvc
  • rvc
second question? how to do?
anonymous
  • anonymous
Like I said, note that \(x_n = e^{i\frac{\pi}{2^{n}}}\) Now multiply the \(x_n\)'s and you will notice an infinite sum in the exponent. Tell me when you get that far.
rvc
  • rvc
@satellite73
Michele_Laino
  • Michele_Laino
please, note that, in the first question, we have not a quadratic equation @hartnn
Michele_Laino
  • Michele_Laino
@nincompoop please help
Michele_Laino
  • Michele_Laino
@dan815 please help
Michele_Laino
  • Michele_Laino
@phi please help
Michele_Laino
  • Michele_Laino
@Loser66 @Hero please help
Michele_Laino
  • Michele_Laino
@mukushla please help
Michele_Laino
  • Michele_Laino
what do you think about the first question? @Hero
anonymous
  • anonymous
sry Michele what's the problem?
Michele_Laino
  • Michele_Laino
if please can help us to solve the first question @mukushla
anonymous
  • anonymous
@rvc please re-check the problem 1, it seems incorrect!
dan815
  • dan815
try just plugging in a and b and get expressions for a and b
rvc
  • rvc
the question is the same
dan815
  • dan815
\[a^2-6\sqrt{a}+4=0\\ b^2-6\sqrt{b}+4=0\]
dan815
  • dan815
not sure where to take it from here u can get some expressions for a^3-b^3
dan815
  • dan815
\[a^3=(-4+6\sqrt{a})^{3/2}\\ b^3=(-4+6\sqrt{b})^{3/2}\] or \[(\frac{(a^2+4)}{6})^6=a^3 \\ (\frac{(b^2+4)}{6})^6=b^3\]
dan815
  • dan815
oh i checked ur solution on wolfram i think your question is wrong, because http://www.wolframalpha.com/input/?i=x^2-6*sqrt%28x%29%2B4%3D0 that equation as real roots
Michele_Laino
  • Michele_Laino
thanks! for your replies @dan815
dan815
  • dan815
yep sure thing :)
rvc
  • rvc
thank you so much :)

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