Question

Find two consecutive positive integers,sum of whose square is 613

Answer 1

let two consecutive positive integers be X, x + 1

according to the question

x² + (x + 1)² = 613

x² + x² + 1 + 2x = 613

2x² + 2x = 613 - 1

2x² + 2x - 612 = 0

2 ( x² + x - 306 ) = 0

x² + x - 306 = 0

x² - 18x + 17x - 306 = 0

x (x - 18) + 17 (x - 18) = 0

(x - 18) (x + 17) = 0

x = 18

x = -17

Hope my answer helps you!!

Answer 2

Answer:

17 and 18

Step-by-step explanation:

Let the smaller of the two consecutive positive integers be x. Then, the second integer will be x + 1.

According to the question,

x² + (x + 1)² = 613

⇒ x² x² + 2 * x * 1 + 1² = 613

⇒ x² + x² + 2x + 1 - 613 = 0

⇒ 2x² + 2x - 612 = 0

⇒ 2 (x² + x - 306) = 0

⇒ x² + x - 306 = 0

Now, we got a quadratic equation,

⇒ x² + x - 306 = 0

⇒ x² + 18x - 17x - 143 = 0

⇒ x ( x + 18 ) - 17 ( x + 18 ) = 0

⇒ ( x - 17 ) ( x + 18 ) = 0

⇒ x = 17 or x = - 18

But x is given to he an odd positive integer. Therefore, x ≠ -18, x = 17.

Thus, the two consecutive odd integers are 17 and 18.