Question

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

Answer 1

Answer:

11 and 13

Step-by-step explanation:

Let a be the first of the two odd integers.  The next odd integer is then a+2.

The sum of their squares = 290

=> a² + (a+2)² = 290

=> a² + a² + 4a + 4 = 290

=> 2a² + 4a + 4 = 290

=> a² + 2a + 2 = 145

=> (a+1)² + 1 = 145

=> (a+1)² = 144

=> a+1 = ±12

=> a = ±12-1

Since we are told that a is positive, it must be

a = 12-1 = 11

and

a+2 = 13

Answer 2

Sol-

Let the first no. be x and the 2nd be (x+2)

According to the given condition,

x^2 + (x+2)^2 = 290

x^2 + x^2 + 4x + 4 = 290

2 x^2 + 4x +4 - 290 = 0

2x^2 +4x - 286 =0

x^2 + 2x - 143 = 0 (Dividing by 2)

x^2 + 13x -11x -143 =0

x (x +13) - 11 (x +13) = 0

(x -11) (x +13) =0

x -11=0 or x + 13 =0

x= 11 or x = -13

But x is a positive integer

x = 11

and x + 2 = 11+2 = 13

Thus, the two nos. are 11 and 13.

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