SOLUTION: Find three consecutive multiples of 5 such that the sum of the square of the first two is 125 greater than the square of the third.

Question 67297This question is from textbook An Increamental Development
: Find three consecutive multiples of 5 such that the sum of the square of the first two is 125 greater than the square of the third. This question is from textbook An Increamental Development

Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
x^2+(x+5)^2=(x+10)+125
x^2+x^2+10x+25=x+135
2x^2+10x-x+10-135=0
2x^2+9x-125=0 using the quadratic equation x=(-b+-sqrt[b^2-4ac])/2a we get
x=(-9+-sqrt[81-4*2*-125])/2*2
x=(-9+-sqrt[81+000])/4
x=(-9+-sqrt1081)/4
x=(-9+-32.87856)/4
x=(-9+32.87856)/4
x=23.878564/4
x=5.96964 solution
x=(-9-32.87856)/4
x=-41.87856/4
x=-10.46964 solution

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