This question is pertaining to college algebra

Question

The sum of the squares of two consecutive even integers is 52. Find the integers.

Answer 1

I1 = 2x; I2 = 2x+2

(2x)² + (2x+2)² = 52

(square the terms)

4x² + 4x² + 8x + 4 = 52

(combine like terms)

8x² + 8x + 4 = 52

(make equation =0)

8x² + 8x -48 = 0

(divide by 8)

x² + x -6 = 0

(factor)

(x+3)(x-2)=0

(only looking for the even integer)

x = 2

(return to fill in)

I1= 4; I2=6

 

Go Get Your "A"

Answer 2

Hello, thank you for taking the time to post your question!

The key on this type of question is to use the underlying information in order to set up an expression that we can solve algebraically. On this one we have

x^2 + (x + 2)^2 = 52

x^2 + x^2 + 4x + 4 = 52

2x^2 + 4x – 48 = 0

x^2 + 2x – 24 = 0

(x + 6)(x – 4) = 0

x = -6 or x = 4

so that means there are actually two pairs of consecutive even integers that work

(-6, -4) and (4, 6)

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 3

Here is another way to solve this problem.  It actually has two solutions:

 

Let the 2 even integers be n and n + 2.

 

Then

 

n² + (n + 2)² = 52

n² + (n² + 4n + 4) = 52

2n² + 4n - 48 = 0

 

Divide through by 2 and get 

 

n² + 2n - 24 = 0 

 

Factor the expression above and get

 

(n + 6)(n - 4) = 0

 

The n = -6 or n = 4.  Note the solutions came out even.  There are two solutions:

 

For n = -6, the two consecutive even integers are -6 and -4.

For n = 4, the two consecutive even integers are 4 and 6.

Since the problem statement did not say the integers had to be positive, both solutions are valid.