find 3 consecutive integers such that the product of the second and third integer is 20
Take three integers x, y, and z.
The for xyz, we want y*z = 20
The factors of 20 are
20*1
10*2
5*4.
20*1 are not consecutive.
10*2 are not consecutive.
But 5 and 4 are consecutive; therefore, we just add 3 in front to make it 345.
These are three consecutive integers and the second and third multiply to give 20.
n, n+1, n+2 are the integers.
(n+1)(n+2)=20
solve for n.
n+1 = 5
n+2 = 4
n = 3
Therefore, the three consecutive integers are 3, 4, and 5.
To solve for n, we expand the quadratic equation as follows:
(n + 1)(n + 2) = 20
Expanding the equation, we get:
n^2 + 3n + 2 = 20
Rearranging the equation, we have:
n^2 + 3n + 2 - 20 = 0
Combining like terms, we get:
n^2 + 3n - 18 = 0
Now, we solve this quadratic equation. We can factorize the equation or use the quadratic formula.
Factorizing:
(n + 6)(n - 3) = 0
Setting each factor to zero and solving for n, we have two possible solutions:
n + 6 = 0 --> n = -6
n - 3 = 0 --> n = 3
Therefore, the two possible values for n are -6 and 3.
If we substitute these values back into the consecutive integer sequence (n, n+1, n+2), we get:
For n = -6: -6, -5, -4
For n = 3: 3, 4, 5
Thus, the two sets of three consecutive integers where the product of the second and third integer is 20 are {-6, -5, -4} and {3, 4, 5}.
To solve for n, we can set up the equation:
(n + 1)(n + 2) = 20
Expanding the equation:
n^2 + 3n + 2 = 20
Rearranging the equation:
n^2 + 3n - 18 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Let's factor the equation:
(n + 6)(n - 3) = 0
Setting each factor equal to zero:
n + 6 = 0 or n - 3 = 0
Solving for n:
n = -6 or n = 3
Therefore, the two possible values for n are -6 and 3.
Using these values, we can find the consecutive integers:
For n=-6:
The three consecutive integers are: -6, -5, -4
For n=3:
The three consecutive integers are: 3, 4, 5