Question

find two consecutive odd positive integers, sum of whose squares is 290

Answer 1

let consecutive no.be x,x+2
square is x^2,x+2^2
x2+x2+4x+4= 290
2x2+4x+4=290
2x2+4x=286
2x2+4x-286=0
2(x2+2x-143)=0
x2+13x-11x-143=0
x(x+13)-11(x+13)=0
x-11=0 x+13=0
x=11
x=-13
therefore x is 11
x=11
x+2=11+2=13
consecutive no.are 11&13
hope this helps you quickly

Answer 2

Solution:

Let two consecutive odd positive integers are x and x + 2.

According to question,

=> x² + (x + 2)² = 290

=> x² + x² + 4x + 4 = 290

=> 2x² + 4x = 290 - 4

=> 2x² + 4x - 286 = 0

Take 2 common from above equation, we get

=> x² + 2x - 143 = 0

By using splitting middle term method we find the value of x.

=> x² + 2x - 143

=> x² - 13x + 11x - 143

=> x(x - 13) + 11(x - 13)

=> (x + 11)(x - 13)

=> x = -11 and 13

∴ Here x is a odd positive integers. So, x = 11 and 13.

Verification:

=> (11)² + (13)² = 290

=> 121 + 169 = 290

=> 290 = 290

LHS = RHS