anonymous
  • anonymous
solve for completing the square. x squared -8x+12=0. show work please :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
*by
lgbasallote
  • lgbasallote
x^2 - 8x +12 = 0 x^2 - 8x = -12 x^2 - 8x +(8/2)^2 = -12 + (8/2)^2 x^2 - 8x + (4)^2 = -12 + (4)^2 x^2 - 8x + 16 = -12 + 16 (x-4)^2 = 4 x-4 = sqrt (4) x - 4 = +/- 2 x = 4 +/- 2 x = 6 x = 2
anonymous
  • anonymous
x squared -8x +12 = 0 -12 -12 x squared -8x + ____ = -12 + ____ (-8/2)squared -4squared 16 x squared -8x + 16 = -12 +16 (x-4)squared=4 (x-4)=+/-2 +4 +4 x=4+/- 2 x=6 x=2

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anonymous
  • anonymous
Ah! Beat me! ;)
lgbasallote
  • lgbasallote
nice game :) you put up a good fight hehe
ash2326
  • ash2326
@ineedhelppleaseee Whenever you need to complete square and you have say \[x^2+bx+c=0\] Just add and subtract \(\frac{b^2}{4}\) We get now \[x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}+c=0\] We'll have now \[(x+\frac{b}{2})^2=\frac{b^2}{4}-c\] Now you can solve by taking square root both sides \[(x+\frac{b}{2})=\sqrt{\frac{b^2}{4}-c}\] so we get \[x=\sqrt{(\frac{b^2}{4}-c})-\frac{b}{2}\] Always remember this and you can solve any quadratic
ash2326
  • ash2326
Sorry you'll get in the last step \[x=\pm\sqrt{(\frac{b^2}{4}-c)}-\frac{b}{2}\]
anonymous
  • anonymous
Hey ash - how do you create the vertical fractions?
ash2326
  • ash2326
lgbasallote
  • lgbasallote
and ash just showed you how to derive the quadratic formula :DDD
anonymous
  • anonymous
As long as a=1, yes. :)
anonymous
  • anonymous
Thanks Ash!

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