three consecutive positive numbers are such that the square of middle number exceeds the difference of the squares of the other two by 60. find the numbers
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let the numbers be x-1, x, x+1
x²-60=(x+1)²-(x-1)²
x²-60=4x
x(x-4)=60=10*6 or x(x-4)=-10*-6
hence x=10 (or -6 but x is positive so -6 is not considered)
numbers are 9,10,11
x²-60=(x+1)²-(x-1)²
x²-60=4x
x(x-4)=60=10*6 or x(x-4)=-10*-6
hence x=10 (or -6 but x is positive so -6 is not considered)
numbers are 9,10,11
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