anonymous
  • anonymous
Solve by completing the square x^2-20x+30=0
Mathematics
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anonymous
  • anonymous
Solve by completing the square x^2-20x+30=0
Mathematics
chestercat
  • chestercat
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KaiserTheSlayer
  • KaiserTheSlayer
using this would help you and remember the a=1, b=-20, and c=30
anonymous
  • anonymous
\[x^2-2*x*10+10^2-10^2+30=0\]
anonymous
  • anonymous
what do you do after that? @surjithayer

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anonymous
  • anonymous
do i used the quadratic formula to solve by completing squares? @KaiserTheSlayer
KaiserTheSlayer
  • KaiserTheSlayer
yes
KaiserTheSlayer
  • KaiserTheSlayer
so it would be x=18.36660026534076 and x= 1.63339973465925
anonymous
  • anonymous
oh okay thank you
KaiserTheSlayer
  • KaiserTheSlayer
yup
ChillOut
  • ChillOut
Solving by square completion yields:\[x²-20x=-30\]\[x²-20x + 100 = 70\]\[(x-10)²=70\] There you have it. Do you the step-by-step?
anonymous
  • anonymous
no i don't
anonymous
  • anonymous
\[(x-10)^2-100+30=0\] \[\left( x-10 \right)^2=70=\left( \sqrt{70} \right)^2\] \[x-10=\pm \sqrt{70}\] x=?
anonymous
  • anonymous
where is the 100 coming from?
ChillOut
  • ChillOut
You take half of the x coefficient, square it and add to both sides.
anonymous
  • anonymous
thats right! i sorta remember now
ChillOut
  • ChillOut
Don't forget to divide everything by whatever is multiplying the squared term ;). In this case, it's "1", so it won't matter.
anonymous
  • anonymous
so then to figure out what x is I solve... x-10=\[\sqrt{70}\] and x-10=\[\sqrt{-70}\]
anonymous
  • anonymous
idk why it looks like that i was trying to write what you did on the last thing
ChillOut
  • ChillOut
Be careful! The second solution you gave is a COMPLEX number.
anonymous
  • anonymous
i know i have to do the whole i stuff for that cause its a negative square root
ChillOut
  • ChillOut
No, it's not:\[x-10=\sqrt{70} \]or\[x-10=-\sqrt{70}\]
ChillOut
  • ChillOut
notice the minus sign is outside.
anonymous
  • anonymous
yeah i see that now
ChillOut
  • ChillOut
And that's it! you have your solutions! Just move the independent term to the other side.
anonymous
  • anonymous
yeah i got it now, thank you!
ChillOut
  • ChillOut
The whole thing about completing the square is to make the left side equation look like something like (x-a)² or (x+a)²

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