The sum of the squares of two positive integers is 208. If the square of the large number is 18 times the smaller. Find the numbers
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let the numbers be x,y (x>y) here x2=18yaccording to the problemx2+y2=208....................(1) 18y+y2=208y2+18y-208=0 y2+26y-8y-208=0y(y+26)-8(y+26)=0 (y+26)(y-8)=0 as the numbers are +ve y-8=0 y=8 substitute y in (1) x2+(8)2=208 x2+64=208 x2=208-64x2=144 x=12 the nos. are 12,8
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Let x be the smaller number.
So, the square of larger number = 18x
A/C,
Square of smaller number + Square of larger number = 208
As x is a positive integer, x ≠ -26 ,so x = 8
Smaller number = 8 and
Smaller number = 8 andLarger number = √18×8 = 12
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