anonymous
  • anonymous
Complete the square: (5/7)x^2+(9/10)x+(-13/70)=0
Mathematics
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anonymous
  • anonymous
Complete the square: (5/7)x^2+(9/10)x+(-13/70)=0
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
ax^2+bx+c
anonymous
  • anonymous
do i multiply by 7/5 to get the coefficient of the squared term to be 1?
anonymous
  • anonymous
ax^2+2ab+b^2

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anonymous
  • anonymous
Divide everything by 5/7
anonymous
  • anonymous
a=5/7 b=9/10
anonymous
  • anonymous
firstly divide everything by 5/7 x^2 + (63/50)x-(-13/50) = 0 (x+63/100)^2 - 3969/10000 - 13/50 = 0 (x+63/100)^2 = 6569/10000 x = - (63/100)+-sqrt(6569/10000)
anonymous
  • anonymous
then evaluate that .. that should hopefully be right (although i did it in my head, method is sound tho)
anonymous
  • anonymous
looking over my numbers...
anonymous
  • anonymous
got it! just went wrong on one of those huge fractions
anonymous
  • anonymous
you are amazing!
anonymous
  • anonymous
alternatively it is usually better instead of dividing it properly by 5/7, to put it outside a large bracket and leave it there, from start; 5/7[x^2+(63/65)x-(13/50) = 0
anonymous
  • anonymous
forgot end bracket .. but u get the gist
anonymous
  • anonymous
yes, great
anonymous
  • anonymous
ok glad i could help
anonymous
  • anonymous
I really appreciate it!

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