anonymous
  • anonymous
Determine the values of 3 consecutive odd integers such that the sum of the squares of the first 2 is 15 less than the 3rd.
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Let x be the integer. x+2 = second odd integer x+4 = third consec. odd integer (x^2)+(x+2)^2= (x+4)-15 When it saaid 15 less than the 3rd, does that mean 15 less than the square of the third integer?
anonymous
  • anonymous
I believe so .. I did the exact same thing .. but i just dont understand what im doing :S
anonymous
  • anonymous
assume the first integer to be (2x+1), the second (2x+3), the third (2x+5)

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anonymous
  • anonymous
\[x^{2}+(x+2)^{2}=(x+4)^{2}-15\] \[2x^{2}+4x+4=x^{2}+8x+1\] \[x^{2}+4x+3=0\] \[(x+1)(x+3)=0\] This is what I got .. but I need three numbers not 2 :S And we have to solve them by using roots ..
anonymous
  • anonymous
the numbers are -1,-3 and -5
anonymous
  • anonymous
??
anonymous
  • anonymous
1,3 and 5 as well
anonymous
  • anonymous
1^2+3^2=5^2-15.. right?
anonymous
  • anonymous
im really confused .. :S

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