Answer:
The first set of consecutive even integers equals (8 , 6)
The second set is ( - 8 and - 6) which also works.
Step-by-step 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.
Answer:
6 and 8
Step-by-step 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.
