The sum of square of two consecutive integer is 613 find them
Answers
ATQ, the sum of two consecutive integers is 613.
consecutive integers are basically adjacent integers. for example :- 1 and 2 are consecutive integers. we just add 1 to the previous number.
let the two consecutive integers be x and (x + 1)
therefore (x)² + (x + 1)² = 613
➡ x² + (x)² + 2(x)(1) + (1)² = 613
➡ x² + x² + 2x + 1 = 613
➡ 2x² + 2x + 1 - 613 = 0
➡ 2x² + 2x - 612 = 0
➡ 2(x² + x - 306) = 0
➡ x² + x - 306 = 0
using splitting the middle term method,
➡ x² + (18x - 17x) - 306 = 0
➡ x² + 18x - 17x - 306 = 0
➡ x(x + 18) - 17(x + 18) = 0
➡ (x + 18) (x - 17)
» x = -18 or x = 17
if x = -18, then It's consecutive integer will be = -18 + 1 = -17
and if x = 17, then it's consecutive integer will be = 17 + 1 = 18
hence, the numbers are either -18 and -17 or 17 and 18
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.
