anonymous
  • anonymous
I need help with the intermediates steps of a sum of squares problem. I have a trinomial and the answers it factors into but cannot work it backwards correctly
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
a sum of squares will have complex numbers in the factorized form
anonymous
  • anonymous
the squar is always equal for example: For the first one: The goal of these problems is to get the left hand side of the equation to be a perfect square and then take the square root of both sides and solve for answers. x^2+16x+64=81 The left side of the equation factors into (x+8)^2 so taking the square root of both sides x+8=+ or - 9 so x= 1 or x= -17
anonymous
  • anonymous
are you talking about factoring a sum of squares or completing the square?

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anonymous
  • anonymous
can i get a medal plzzzzzzzzzzzzzzzzzzz
anonymous
  • anonymous
if y^2-2xy-x^2=0 and the answers are y=(-1+,- 2^.5)x what are the intermediate steps working backward from those answers?
anonymous
  • anonymous
you have to try to get the left side of the equation to a perfect square then you square root both sides then solve
anonymous
  • anonymous
medal plzzz
anonymous
  • anonymous
sum of squares! \[f(1)=1^2+f(0)\] \[f(2)=2^2+f(1)\] . . . \[f(n)=n^2+(f(n-1))\] Sum all the expressions above: \[\sum_{1}^{n}(x^n)=f(n)-f(0)\] \[f(n)=ax^3+bx^2+cx+d\] \[\sum_{1}^{n}(x^n)=ax^3+bx^2+cx+d\] this is the notion you need. then you compare polynomials and get to the answer. i think it is \[f(n)=(n^3)/6+(n^2)/2+n/3\]
anonymous
  • anonymous
just dunno if I get the question right! hehe
anonymous
  • anonymous
thanks both of you but I need to work backward from the answer
anonymous
  • anonymous
to rearrive at y^2-2xy-x^2=0?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I think I am totally off topic haha, can't wait to see the answer
anonymous
  • anonymous
\[y^2-2xy-x^2=0\] \[(y-x)^2 -2x^2 = 0\] \[y - x = \pm (\sqrt{2})x\] \[y = x \pm (\sqrt{2})x = (1 \pm \sqrt{2})x\]
anonymous
  • anonymous
thanks but I'm timed out on my library computer
anonymous
  • anonymous
ahh
anonymous
  • anonymous
y^2−2xy−x^2=0 (y−x)^2 − 2x^2=0 y−x=±(2^0.5)x y=x±(2^0.5)x=(1±(2^0.5))x
anonymous
  • anonymous
i think some people got confused when you called it "sum of squares" instead of "completing the square" :)
anonymous
  • anonymous
this works as y^2 - 2xy -x^2 = y^2 - 2xy + x^2 -2x^2 = (y-x)^2 - 2x^2

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