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one positive number is two more than another. The sum of there squares is 34. Find the smaller number
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one number is x, the other is x +2. x + (x+2)^2 = 34 solve for x
its actually \[x^{2}+(x+2)^{2}=34\]
yepp...typo error
expand (x+2)^2
2x^2 +4x + 4 = 34
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or x ^2 +2x -15 = 0
(x +5 )(x - 3) = 0
\[x^2+(x^2+4x+4)=34\]now combine like terms\[2x^2+4x+4=34\]now move the 34 over and factor to solve for x\[2x^2+4x-30=0\]
x = -5 or x = 3
sandy is correct! :)
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thanks!
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