Question

find 2 consecutive positive integer sum of whose square is 221

Answer 1

Hii friend,

Let the first consecutive positive integer be X .

Then the other consecutive positive will be (X+1).

According to question,

(X+1)² + (X)² = 221

X²+(1)² + 2 × X × 1 + X² = 221

X²+1+2X+X² = 221

2X²+2X-220 = 0

2(X²+X-110) = 0

X²+X-110 = 0

X²+11X-10X -110 = 0

X(X+11) - 10(X+11) = 0

(X+11) (X-10) = 0

(X+11) = 0. OR (X-10) = 0

X = -11 OR X = 10

Hence,

The two consecutive positive integer are 10 and 11.


HOPE IT WILL HELP YOU.... :-)