What is the smallest of 3 consecutive positive integers if the product of the smallest and largest is 21 more than 9 times the middle number?
x(x+2)=21+9(x+1)
I'm not sure if this is a good equation to solve it..?
They are POSITIVE, odd numbers are OUT
so they differ by TWO
n-2 , n , n+2
(n-2)(n+2) = 21 + 9 n
n^2 - 4 = 21 + 9 n
n^2 - 9 n -21 = 0
(n-7)(n+3) = 0
n = 7
so the smallest is 5
You're on the right track! We can set up an equation to solve this problem.
Let's assume that the smallest of the three consecutive positive integers is represented by the variable x.
According to the given information, the middle number would be represented by (x + 1), and the largest number would be represented by (x + 2).
The problem states that the product of the smallest and largest numbers is 21 more than 9 times the middle number. Mathematically, we can write this as:
x * (x + 2) = 9 * (x + 1) + 21
Now, we can solve this equation to find the value of x, which represents the smallest of the three consecutive positive integers.
Multiplying out the left side of the equation gives us:
x^2 + 2x = 9x + 9 + 21
Simplifying further:
x^2 + 2x = 9x + 30
Rearranging the equation:
x^2 + 2x - 9x - 30 = 0
Combining like terms:
x^2 - 7x - 30 = 0
Now, we have a quadratic equation that we can solve using factoring, completing the square, or quadratic formula. In this case, let's use factoring:
(x - 10)(x + 3) = 0
Setting each factor equal to zero:
x - 10 = 0 or x + 3 = 0
Solving for x in each equation:
x = 10 or x = -3
Since we are looking for the smallest positive integer, we don't consider the negative solution. Therefore, the smallest of the three consecutive positive integers is 10.
So, the answer is 10.