Question

find two consecutive positive integers, sum of whose squares is 365

Answer 1

let the first no be x and second be x+1.
a/q
(x)^2 + (x+1)^2 = 365
x^2+x^2+1+2x = 365
2x^2+2x-364 = 0
2x^2+(28-26)x-364 = 0
2x^2+28x-26x-364 = 0
2x(x+28)-26(x+28) = 0
(x+28)(2x-26) = 0
x = 26/2 = 13; x = -28(not possible)
first no = 13
second no = 14

Answer 2

Heya!!!

Let the consecutive integers be

'x' and 'x+1'

Given : x^2 + (x+1)^2

= x^2 +x^2 + 2x +1 = 365

= 2x^2 + 2x - 364 = 0

= x2+x−182=0

=Let's solve your equation step-by-step.

x2+x−182=0

Step 1: Use quadratic formula with a=1, b=1, c=-182.

x=−b±√b2−4ac2a

x=−(1)±√(1)2−4(1)(−182)2(1)

x=−1±√7292

x=13 or x=−14

Since its a positive integer , -14 would b rejected.

Thus the two consecutive positive integers are

13 and 14

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