|
Question 209804: find three consecutive positive integers such as that the square of the first, increased by the last is 22
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Consecutive integers are one apart from each other. So if the smallest one is "x", then the next two will be "x+1" and "x+1+1". So the statement "the square of the first, increased by the last is 22" translates into:
To solve this we will start by simplifying the left:
Then, because this is a quadratic equation, we will get one side equal to zero (by subtracting 22 from both sides):
Now we can solve by factoring (or, if you prefer, the quadratic formula):
In order for this product to be zero, one of the factors must be zero:
 or 
Solving these two we get:
 or 
Since the problem stipulates positive integers, we must exclude the first solution. So the three consecutive positive integers (x, x+1 and x+1+1) are: 4, 4+1, 4+1+1 or 4, 5, 6.
|
|
|
| |
