How do i get 2 consecutive numbers (pair numbers) that they product is 272?

4X² + 4X - 272=0

h, h +1

h(h+1) = 272

h^2 + h = 272

h^2 + h -272 = 272-272

h^2 + h -272 = 0

(h -16)(h + 17) = 0

h = 16
h = -17

-16, -17, and 16, 17

x = first number

y = second number

y = x + 1

x * y = 272

x * ( x + 1 ) = 272

x ^ 2 + x = 272

x ^ 2 + x - 272 = 0

Solutions :

x = - 17

and

x = 16

For x = - 17

y = x + 1 = - 17 + 1 = - 16

x * y = - 17 * ( - 16 ) = 272

For x = 16

y = x + 1 = 16 + 1 = 17

x * y = 16 * 17 = 272

So solutions are :

- 17 and - 16

and

16 and 17

Yes.Both are all solution.

Proof

x(x+1)= 272

16(16+1) = 272
16(17) =272
272 = 272

-17(-17+1) = 272

-17(-16)= 272
272 = 272

To find a pair of consecutive numbers whose product is 272, you can start by setting up a quadratic equation. Let's assume that the first number is represented by "X" and the second consecutive number is "X + 1".

To solve this problem, you need to find the values of X that satisfy the equation 4X² + 4X - 272 = 0.

To simplify the equation, you can divide each term by 4:
X² + X - 68 = 0

Now, we can use the quadratic formula to solve for X:
X = (-b ± √(b² - 4ac)) / 2a

Substituting the values from the equation, we have:
X = (-(1) ± √((1)² - 4(1)(-68))) / 2(1)
X = (-1 ± √(1 + 272)) / 2
X = (-1 ± √273) / 2

Since we are looking for two consecutive integers, we can evaluate the expression using the positive square root (√273) first:
X = (-1 + √273) / 2 ≈ 8.196

Since X should be an integer, this value doesn't work.

Next, evaluate the expression using the negative square root (-√273):
X = (-1 - √273) / 2 ≈ -9.196

Again, since X should be an integer, this value doesn't work either.

Thus, there are no two consecutive numbers whose product is 272.