Question

Find the consecutive positive integer sum of whose squares is 365

Answer 1

Answer:

ler the two consecutive integers be X and y

Then according to the question

x square +(x+1)square =365

= X square +X square + 2x+1 - 365=0

=2x square +2x -364=0

= take 2 common

=2 ( X square +X -182) = 0

X square + X - 182 =0

X square +14x -13x - 182 =0

(x+14) (x-13)=0

x=-14 or x=13

Answer 2

Given:-

• Sum of whose is 365.

To find:-

• Two consecutive positive integer...?

Solutions:-

• Let the conservative 1st position integer be x.
• Let the conservative 2n position integer be x + 1.

we know that;

=> x² + (x + 1)² = 365

=> x² + x² + 1 + 2x = 365

=> 2x² + 2x - 364 = 0

=> x² + x - 182 = 0

=> x² + 14x - 13x - 182 = 0

=> x(x + 14) - 13(x + 14) = 0

=> (x + 14) (x - 13) = 0

=> x + 14 = 0 or x - 13 = 0

=> x = - 14 or x = 13

Since, the integers are positive, x can only be 13.

Therefore,

=> x + 1

=> 13 + 1

=> 14

Hence, the consecutive positive integer be 13 and 14.