Find two consecutive even integers whose sum squared is 100.

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- Dec 3rd 2008, 05:17 AM #1

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- Dec 3rd 2008, 06:10 AM #2

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Find two consecutive even integers whose sum squared is 100.

Now my two numbers will be x and x+2 and their sum -squared- will be 100.

$\displaystyle (x+x+2)^2=100$

$\displaystyle (x+x+2)(x+x+2)=100$

$\displaystyle x^2+x^2+2x+x^2+x^2+2x+2x+2x+4=100$

$\displaystyle 4x^2+4x+4-100=0$

$\displaystyle 4x^2+4x-96=0$

$\displaystyle 4x^2+4x-96=0$ /4

$\displaystyle x^2+x-24=0$

Now I try to find the factor for middle term

$\displaystyle 1(-24)=-24$ $\displaystyle -23$

$\displaystyle 2(-12)=-24$ $\displaystyle -10$

$\displaystyle 3(-8)=-24$ $\displaystyle -5$

$\displaystyle 4(-6)=-24$ $\displaystyle -2$

$\displaystyle 6(-4)=-24$ $\displaystyle +2$

Mmm...I Can't find two factors that add to +1...Grrr....

- Dec 3rd 2008, 08:28 AM #3

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- Dec 3rd 2008, 08:38 AM #4

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