Find three consecutive positive integers such that the product of the first and third interger is 17 more than 3 times the second interger

5, 6, 7

Step-by-step explanation:

In order to solve for the three integers, we can assign a variable and set up an equation:

first integer: x

second integer: x + 1

third integer: x + 2

Given that 'the product of the first and third integer is 17 more than 3 times the second integer':

x(x + 2) = 3(x + 1) + 17

Distribute: x² + 2x = 3x + 3 + 17

Combine like terms: x² - x - 20 = 0

Factor: (x - 5)(x + 4) = 0

Set them equal to '0' and solve:

x - 5 = 0      x + 4 = 0

x = 5               x = -4

Since the problem asks for positive integers, x must equal 5:

first = 5

second = 5 + 1 = 6

third = 5 + 2 = 7

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length