Question

if the sum of the squares of two consecutive even positive integer is increased by 100 then it will be 10 times the difference of their square find the integer​

Answer 1

The consecutive positive integers are (2, 4) and (16 ,18)  

Step-by-step explanation:

Given as :

Let The consecutive even integers = 2 x ,  2 (x + 1)

Square of integers = (2 x)²   , [ 2 (x + 1) ]²

According to question

Statement I

The sum of the squares of two consecutive even positive integer is increased by 100

i.e   [ 2 (x + 1) ]² +  (2 x)² + 100              ........1

Statement II

10 times the difference of their square find the integer​

So,  10 × [ ( 2 (x + 1) )²  -   (2 x)² ]                 ............2

From eq 1 and eq 2

  [ 2 (x + 1) ]² +  (2 x)² + 100  =  10 × [ ( 2 (x + 1) )²  -   (2 x)² ]

Or, (2 x)² + 4 + 8 x +  (2 x)² + 100  =  10 × [  (2 x)² + 4 + 8 x  -   (2 x)² ]

Or,  (2 x)² + 4 + 8 x +  (2 x)² + 100  =  10 ×  ( 4 + 8 x )

Or,  (2 x)² + 4 + 8 x +  (2 x)² + 100  =  40 + 80 x

Or, 8 x² + 8 x - 80 x + 104 - 40 = 0

Or,   x² +  x - 10 x + 13 - 5 = 0

Or, x²  - 9 x + 8 = 0

Or,  x² - 8 x - x + 8 = 0

Or,  x ( x - 8 ) - 1 ( x - 8 ) = 0

Or, ( x - 8 ) ( x - 1 ) = 0

∴  ( x - 8 ) = 0    ( x - 1 ) = 0

i.e x = 8  , 1

So, The consecutive even integers = 2 × 1 = 2

And                                                     = 2 (1 + 1) = 4

Similarly

The consecutive even integers = 2 × 8 = 16

And                                                     = 2 (8 + 1) = 18

Hence, The consecutive positive integers are (2, 4) and (16 ,18)   Answer

Answer 2

1 st term = 2   and  2nd term = 4

Given:

Sum of the squares of two consecutive even positive integer is increased by 100 then it will be 10 times the difference of their square.

Find:

The integer.

Explanation:

There are two consecutive even  positive integer.

Consider the 1st value = x

                    2nd value= x+2

From given information

                 $x^{2}$ + $(x+2)^{2}$ +100 = 10×  ($(x+2)^{2}$- $x^{2}$ )

To solve this equation this will take more time so we can solve this by trial and error

so initially we can stat x=2 because even no

            $2^{2}$ + $(2+2)^{2}$ +100 = 10×  ($(2+2)^{2}$ - $2^{2}$ )

             4+4² +100 = 10× (4² - 4)

                   120 = 120

So the equation can be satisfied

1 st term = x = 2

2nd term = x+2 = 2+2 = 4

To learn more...

1 Why do we reduce the expression with the help of boolean algebra and demorgan's theorem

https://brainly.in/question/11506128

2 The algebra of sum of an arithmetic sequence is [n]2+2n. a.) What is the sum of first 10 term of this sequence. b.) How many terms from the first ate needed to get the sum 168?

https://brainly.in/question/6161260