# Math Help - Let's see if any of you have the same

1. ## Let's see if any of you have the same

Find two consecutive even integers whose sum squared is 100.

2. 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.

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

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

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

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

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

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

$x^2+x-24=0$

Now I try to find the factor for middle term

$1(-24)=-24$ $-23$
$2(-12)=-24$ $-10$
$3(-8)=-24$ $-5$
$4(-6)=-24$ $-2$
$6(-4)=-24$ $+2$

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

$4x^2+8x+4-100=0$

$4x^2+8x-96=0$

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

4. Originally Posted by Alienis Back
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.

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

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

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

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

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

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

$x^2+x-24=0$

Now I try to find the factor for middle term

$1(-24)=-24$ $-23$
$2(-12)=-24$ $-10$
$3(-8)=-24$ $-5$
$4(-6)=-24$ $-2$
$6(-4)=-24$ $+2$

Mmm...I Can't find two factors that add to +1...Grrr....
You could've avoided all this mess by simply taking the square root:

$(2x+2)^2=10^2 \implies 2x+2=\pm 10 \implies x = -1 \pm 5$