Sum of the square of two cojucative natural number is 313 find the numbers
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Let x and x+1 are two consecutive terms
x^2+(x+1)^2=313
x^2+x^2+2x+1-313=0
2x^2+2x-312=0
x^2+x-156=0
x^2+13x-12x-156=0
x(x+13)-12(x+13)=0
(x+13)(x-12)=0
(x+13)=0 or x-12=0
x = -13 or x=12
Required numbers are
12 and 13 or (-13 and -12)
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