Question

Find two consecutive odd positive integers, sum of whose squares is 290
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Answer 1

To find:-

• The two consecutive odd positive integers.

Solution :-

Let the 1st positive odd integer be "x"

and the 2nd positive integer be " x+2"

Given

Sum of two consecutive odd positive integers is 290.

Sum of their squares will be __

x²+ (x+2)²

A.T.Q :-

» x²+ (x+2)² = 290

» x² + x² + 4x + 4 = 290

» 2x² + 4x + 4 = 290

» 2x² + 4x = 286

» 2x² + 4x - 286 = 0

» x² + 2x - 143 = 0

» x² + 13x - 11x - 143 = 0

» x ( x+13) - 11 ( x+13) = 0

» (x-11) (x+13) = 0

Hence,

» x - 11 = 0

» x = 11

OR,

______________________

» x + 13 = 0

» x = -13

Therefore

________

1st positive odd integer is = 11

∴ 2nd positive odd integer will be = (x+2)

= 11 + 2 = 13

________________________________

Answer 2

Let the two odd positive integers are x+1, x+3

According to question :-

=> (x+1)² + (x+3)² = 290

=> x² + 2x + 1 +x² + 6x + 9 = 290

=> 2x² + 8x + 10 - 290 = 0

=> 2x² + 8x - 280 = 0

=> x² + 4x - 140 = 0

=> x² + (14-10)x - 140 = 0

=> x² + 14x -10x - 140 = 0

=> x(x+14) -10(x+14) = 0

=> (x+14)(x-10) = 0

Now, x = -14 and x = 10

from here, x = -14 is rejected

Hence, x = 10

Thus, the odd positive integers are = 11 and 13