Pre Algebra

Question

Find two consecutive positive integers such that the square of the first decreased by 25 equals three times the second.

Answer 1

With this question, you want to write out an equation.

 

Two consecutive positive integers means the numbers are right next to each other, so let's call the first one x and the second one x+1.

 

So, following the word problem, the equation would be: x² - 25 = 3(x+1).

 

Using distribution, we get x² - 25 = 3x + 3

 

If we bring everything over to the left side, we end up with: x² - 3x - 28 = 0.

 

Once we factor this out, we (x-7)(x+4) = 0.

 

That means x has to be either 7 or -4.

 

Since the question asks for a positive integer, x = 7, and the second integer is 8.

 

To verify, we can plug this into the original equation x² - 25 = 3(x+1).

 

7² -25 = 49-25 = 24.

 

3(7+1) = 3*8 = 24.

 

Therefore, the two integers you are looking for are 7 and 8.

Answer 2

Consider this, 

 

if we let the first number be represented by a variable "n", then the second consecutive number could be characterized by n+1.

 

so, the square of the first decreased by 25, n² - 25 equals three times the second 3(n+1). putting this all together

 

n² - 25 = 3(n+1) - use the distributive property on the right side and then set the entire equation equal to zero.

 

n² - 3n -28 = 0 which when factored will yield solutions of either 7 or -4. We can eliminate the negative solution because of our initial conditions. Out final results are the two numbers 7 and 8.