<p>when each side of a given square is lengthened by 2 inches, the area is incresed by 40 square inches. what is the length in inches of a side of the originial square? how do i do this?</p>
<p>assume the length of the side is X.</p>
<p>the area would be X (raised to the 2nd power)
That is area 1</p>
<p>the new length is X + 2</p>
<p>the area would be: (X + 2) raised to the 2nd power
That is area 2</p>
<p>The diff is 40 so</p>
<p>{(x+2)power 2} - {x power 2} = 40</p>
<p>slove it on your own fron there; it's impossible to type it</p>
<p>suck on pretty much explained it but you can also use guess and check. Guess and check would work real well if the answers are already given to you(like the sats) and u can just plug in to c which one works.</p>
<p>By the way: it's sTuck on, not suck on</p>
<p>A quick way to solve (x+2)^2 - x^2 = 40</p>
<p>is to use the formula A^2 - B^2 = (A+B)(A-B) where A=x+2, B=x</p>
<p>From this, we get (x+2+x) (x+2-x) = (2x+2)(2) = 4(x+1) = 40
So x+1 = 40/4 = 10
x = 9</p>
<p>Check: 9^2 = 81, (9+2)^2 = 121, and 121 - 81 = 40 .</p>
<p>The area for a square is side squared, so the equation for the area of the first square we'll call s^2=a</p>
<p>The side of the new square would be (s+2)^2 since the side of the new square is 2 more than the side of the original square, s. You're given the new area is 40 more or a+40. So you have (s+2)^2=a+40. Foil out and you get s^2+4s+4=a+40. You know from the first equation that s^2=a, so a+4s+4=a+40. Subtract a and 4 from both sides to get 4s=36, s=9.</p>
<p>my bad typo</p>