The product of two consecutive integer is 5 three times the larger integer, what are the integers?
To solve this problem, let's start by assigning variables to the integers. Let's call the smaller integer "x" and the larger integer "x + 1" (since they are consecutive).
Now we can set up an equation based on the given information. The product of the two consecutive integers is 5 three times the larger integer (3 times (x + 1)).
So, our equation is:
x * (x + 1) = 5 * 3 * (x + 1)
Next, we can simplify and solve the equation:
x^2 + x = 15x + 15
Rearranging the equation and combining like terms, we get:
x^2 - 14x - 15 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Factoring may not work here since the factors of -15 (the constant term) that add up to -14 (the coefficient of the linear term) cannot be found easily.
Using the quadratic formula, which states that the solutions for the equation ax^2 + bx + c = 0 are given by:
x = (-b ± √(b^2 - 4ac)) / (2a),
we can find the values of x. In this case, a = 1, b = -14, and c = -15.
Plugging these values into the formula gives us:
x = (14 ± √((-14)^2 - 4*1*(-15))) / (2*1)
Simplifying further:
x = (14 ± √(196 + 60)) / 2
x = (14 ± √256) / 2
Now, we can take the square root of 256, which is 16, and simplify:
x = (14 ± 16) / 2
This gives us two possible values for x:
1. x = (14 + 16) / 2 = 15
2. x = (14 - 16) / 2 = -1
So, the two consecutive integers are either 15 and 16 or -1 and 0.