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The difference of the squares of two positive consecutive even integers is 68 find the integers use the fact that if X represents an even integer then X +2 represents the next consecutive even integers

Respuesta :

Answer:

16 and 18

Step-by-step explanation:

Here, x represents the even integer then the even consecutive next integer is x + 2,

According to the question,

[tex](x+2)^2-x^2=68[/tex]

[tex]x^2+4x+4-x^2=68[/tex]  ( Using (a+b)² = a² + 2ab + b² )

[tex]4x+4=68[/tex]                 ( Combining the like terms )

[tex]4x=64[/tex]

[tex]\implies x = \frac{64}{4}=16[/tex]

Hence, first integer = 16,

Second integer = 16 + 2 = 18.

The two numbers should be 16 and 18.

Calculation of integers:

Since x represents the even integer so the even consecutive next integer is x + 2.

Now

[tex](x + 2)^2 - x^2 = 68\\\\x^2 + 4x+ 4 -x^2 = 68[/tex]

4x + 4 = 68

4x = 64

x = 16

So, the second one should be

= 16 + 2

= 18

learn more about integer here: https://brainly.com/question/1506648

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