Solve algebraically using one variable: Find three consecutive odd integers such that the
product of the first integer and the third integer is equal to nine more than twelve times the middle integer.
Every odd integer can be written in the form 2 k + 1
So your first integer is a1 = 2 k + 1
Second integer is a2 = first integer + 2 = a1 + 2 = 2 k + 1 + 2 = 2 k + 3
Third integer is a3 = second integer + 2 = a2 + 2 = 2 k + 3 + 2 = 2 k + 5
Product of the first integer and the third integer is equal to nine more than twelve times the middle integer mean:
a1 * a3 = 12 * a2 + 9
Replace a1 = 2 k + 1 , a2 = 2 k + 3 and a3 = 2 k + 5 in this equation.
( 2 k + 1 ) * ( 2 k + 5 ) = 12 * ( 2 k + 3 ) + 9
2 k * 2 k + 2 k * 1 + 5 * 2 k + 5 * 1 = 12 * 2 k + 12 * 3 + 9
4 k ^ 2 + 2 k + 10 k + 5 = 24 k + 36 + 9
4 k ^ 2 + 12 k + 5 = 24 k + 45 Subtract 24 k to both sides
4 k ^ 2 + 12 k + 5 - 24 k = 24 k + 45 - 24 k
4 k ^ 2 - 12 k + 5 = 45 Subtract 45 to both sides
4 k ^ 2 - 12 k + 5 - 45 = 45 - 45
4 k ^ 2 - 12 k - 40 = 0 Divide both sides by 4
k ^ 2 - 3 k - 10 = 0
Try to solve this quadratic equation.
The solutions are:
k = - 2 and k = 5
You have two set of the solutions:
1)
k = - 2
a1 = 2 k + 1 = 2 * ( - 2 ) + 1 = - 4 + 1 = - 3
a2 = a1 + 2 = - 3 + 2 = - 1
a3 = a2 + 2 = - 1 + 2 = 1
- 3 , - 1 , 1
2)
k = 5
a1 = 2 k + 1 = 2 * 5 + 1 = 10 + 1 = 11
a2 = a1 + 2 = 11 + 2 = 13
a3 = a2 + 2 = 13 + 2 = 15
11 , 13 , 15
To solve this problem algebraically, let's consider the three consecutive odd integers.
Let's assume that the first odd integer is represented by "x", the second odd integer is represented by "x + 2" (since consecutive odd integers have a difference of 2), and the third odd integer is represented by "x + 4" (again, the difference between consecutive odd integers is 2).
According to the given information, the product of the first integer and the third integer is equal to nine more than twelve times the middle integer. We can express this algebraically as:
x * (x + 4) = 12(x + 2) + 9
Now, we can solve this equation algebraically:
x^2 + 4x = 12x + 24 + 9
x^2 + 4x = 12x + 33
We need to move all terms to one side of the equation to solve for x:
x^2 + 4x - 12x = 33
Simplifying the equation:
x^2 - 8x = 33
Bringing all terms to one side:
x^2 - 8x - 33 = 0
Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -8, and c = -33.
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, factoring might not be straightforward, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
x = (-(-8) ± √((-8)^2 - 4(1)(-33))) / (2(1))
x = (8 ± √(64 + 132)) / 2
x = (8 ± √196) / 2
x = (8 ± 14) / 2
This gives us two possible values for x:
1) x = (8 + 14) / 2 = 22 / 2 = 11
2) x = (8 - 14) / 2 = -6 / 2 = -3
Now we know the possible values for the first odd integer. We can find the other two consecutive odd integers by adding 2 and 4, respectively.
For the first case (x = 11):
First odd integer: x = 11
Second odd integer: x + 2 = 11 + 2 = 13
Third odd integer: x + 4 = 11 + 4 = 15
For the second case (x = -3):
First odd integer: x = -3
Second odd integer: x + 2 = -3 + 2 = -1
Third odd integer: x + 4 = -3 + 4 = 1
So, the two sets of three consecutive odd integers that satisfy the given conditions are:
1) 11, 13, 15
2) -3, -1, 1