Question

find two consecutive possitive even integers the sum of whose square is 164

Answer 1

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Answer 2

Two consecutive positive even integers is ,

-1 + $\sqrt{(79)}$ and

1 + $\sqrt{(79)}$

Given:

The number 164

To find:

Two consecutive positive number.

Solution:

Let  two consecutive positive even integers  be x and (x + 2).

We know from the question,

$x^{2}$ + $(x + 2)^{2}$= 164

⇒$2x^{2}$ + 4x - 156 = 0

⇒$x^{2}$ + 2x - 78 = 0

We can solve this quadratic equation using the quadratic formula:

x = [-b ±$\sqrt{(b^2 - 4ac)] / 2a}$

here a = 1,

b = 2,

c = -78.

Now putting this value,

x = [-2 ± $\sqrt{(2^2 - 4(1)(-78))] / 2(1)} }$

⇒x = [-2 ±$\sqrt{(316)]}$ / 2

⇒x = [-2 ± $2\sqrt{(79}$)] / 2

⇒x = -1 ± $\sqrt{(79)}$

If we take positive integer only,

x = -1 + $\sqrt{(79)}$

Other consecutive even integer will be ,

x + 2 = -1 + $\sqrt{(79)}$) + 2

        =1 + $\sqrt{(79)}$

       

So, two consecutive positive even integers is ,

-1 + $\sqrt{(79)}$ and

1 + $\sqrt{(79)}$

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