Question

The sum of squares of two consecutive positive even integers is 52 more than their product. Find these numbers.

Answer 1

6 and 8

Explanation:

Let x and x+2 are two even integers.

From question statement,we noticed that

x²+(x+2)² = x(x+2) +52

As (x+2)²= x²+4x+4

x²+x²+4x+4 = x²+2x+52

adding -x²,-2x and -52 to both sides of above equation,we get

x²+x²+4x+4 -x²-2x-52 = x²+2x+52 -x²-2x-52

add like terms

x²+2x-48 = 0

split the middle term of above equation so that the sum of two terms should be 2 and their product be -48.

x²+8x-6x-48 = 0

make two groups

x(x+8)-6(x+8) = 0

taking (x+8) as common,we get

(x+8)(x-6) = 0

Applying Zero-Product Property to both sides of above equation,we get

x+8 = 0 or x-6 = 0

If x+8 = 0

adding -8 to both sides of above equation,we get

x+8-8 = 0-8

x = -8 which is not possible because -8 is not positive.

If x-6 = 0

adding 6 to both sides of above equation,we get

x-6+6 = 0+6

x = 6

x+2 = 6+2

x+2 = 8

Hence, the two consecutive positive even integers are 6 and 8.

Answer 2

The first set of consecutive even integers equals (8 , 6)

The second set is ( - 8 and - 6) which also works.

Explanation:

Equation

(x)^2 + (x + 2)^2 = (x)(x + 2) + 52 Remove the brackets on both sides

Solution

x^2 + x^2 + 4x + 4 = x^2 + 2x + 52 Collect the like terms on the left

2x^2+ 4x+ 4 = x^2 + 2x + 52 Subtract right side from left

2x^2 - x^2 + 4x - 2x + 4 - 52 = 0 Collect the like terms

x^2 + 2x - 48 = 0 Factor

(x + 8)(x - 6) = 0

Answer

Try the one you know works.

x - 6 = 0

x = 6

Therefore the two integers are 6 and 8

6^2 + 8^2 = 100

6*8 + 52 = 100

So 6 and 8 is one set of consecutive even numbers that works.

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What about the other set.

x + 8 = 0

x = - 8

x and x + 2

- 8 and -8 + 2 = - 8, - 6

(- 8 )^2 + (- 6)^2 = 100

(-8)(-6) + 52 = 100

Both sets of consecutive numbers work.