Q. 135.0( 4 Votes )

The sum of the squares of two consecutive natural numbers is 421. Find the numbers.

Answer :

Let the first number be ‘X’, so the other number will be ’(X+1)’.

X2 + (X + 1)2 = 421

X2 + X2 + 1 + 2X = 421

On simplifying further,

2X2+ 2X – 420 = 0

X2 + X – 210 = 0

On applying Sreedhracharya formula

X = –15 or X = 14.

X = 14 (Only natural number, as given in the question)

Then other numbers will be 14 & 15.

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