Question

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

Answer 1

Let the first positive odd integer be x

the other odd positive integer be x+2

sum of their squares ≈290

⇒x²+[x+2]²≈290

  x²+x²+4x+4≈290

   2x²+4x+4≈290

   2x²+4x≈290-4

  2x²+4x≈286

  This can be written as 2x²+4x-286≈0

⇒2x²+4x-286≈0

this can be written as x²+2x-143≈0[whole equation divide by 2]

⇒x²+13x-11x-143≈0

 x[x+13] - 11[x+13]≈0

[x-11]  [x+13] ≈ 0

x-11 ≈ 0

x ≈ 0+11

x ≈ 11

x+ 13 ≈ 0

x ≈ 0-13

x ≈ -13

⇒ x≈11 or x≈-13

given that odd positive integer ,so we take the positive value of x

x ≈ 11

∴[x+2] ≈ 11+2 ≈ 13

Therefore, odd positive integers are 11 and 13