anonymous
  • anonymous
Find conman solution algebraically and express itrrational roots in simplest radical form. x^2+y^2=24 x+y=8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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shubhamsrg
  • shubhamsrg
where are you stuck ?
anonymous
  • anonymous
I dont know anything at all to be honest.
anonymous
  • anonymous
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anonymous
  • anonymous
You could start with thes second equation and find an expression for x= (or y= if you choose) and then substitute that into the first equation.
anonymous
  • anonymous
I dont get anything =/ second all Just go to shift x to other side right and makes t y=x-8
anonymous
  • anonymous
y=8-x
anonymous
  • anonymous
x^2+(8-x)^2=24
anonymous
  • anonymous
x^2+64-16x+x^2=24
anonymous
  • anonymous
2x^2-16x+40=0
anonymous
  • anonymous
x^2-8x+20=0
anonymous
  • anonymous
so y cnt be x-8 but has to 8-x?
anonymous
  • anonymous
yes..
anonymous
  • anonymous
(x-10)(x+2)=0 x=10 x=-2
anonymous
  • anonymous
so those will be the solutions right?
anonymous
  • anonymous
find b's also..
anonymous
  • anonymous
I meant y
anonymous
  • anonymous
because these eq. are two variable..
anonymous
  • anonymous
could you just guide me through like what to do?
anonymous
  • anonymous
choose one of the eq. it does not matter which one, x+y=8 is easier one than put for x=10 find y for x=-2 find y
anonymous
  • anonymous
thank you!
anonymous
  • anonymous
hold on I made a mistake..
anonymous
  • anonymous
until this it is true, but I made a mistake when I factored it out.. x^2-8x+20=0
anonymous
  • anonymous
ax^2+bx+c=0 d=b^2-4ac d=8^2-4*1*20=-16<0 no real solution..
anonymous
  • anonymous
it means that they are not crossing each other..
anonymous
  • anonymous
sorry about that..
anonymous
  • anonymous
root are complex number..
anonymous
  • anonymous
its okay thanks for your time, really appreciate all this walk through.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=solve+x^2%2By^2%3D24%2Cx%2By%3D8
anonymous
  • anonymous
you can see it here
anonymous
  • anonymous
yes thank you!
anonymous
  • anonymous
np

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