1.

Find two consecutive positive integers ,sum of whose square is 365

Answer» Let us two positive integer be x and x+1A.T.Qx^2+(x+1)^2=365x^2+x^2+2x +1=3652x^2+2x-364=0x^2+x-182=0Let us split middle termX^2+14x-13x-182=0 X(x+14)-13 (x+14)=0(X+14)(X-13)=0X =-14 or X=13Negative root being rejectedHence required number are 13 and 14
13 and 14


Discussion

No Comment Found