I need algebra help.?

This is a perfect square...
x²y² - 81
x²y² - 9²
(xy + 9)(xy - 9)

n + n² = 56

n + n² - 56 = 0
n² + n - 56 = 0
(n + 8)( n - 7) = 0

n = -8 or 7

for the second one:
56 = x + x^2
0 = x^2 + x - 56
use the quad formula: (-b +/- sqrt(b^2 - 4ac)) / 2a
x = 7 or -8

#1...(xy + 9) * (xy - 9)
#2... x^2 + x = 56
this can be written as ... x^2 + x - 56 = 0
factoring, we will get (x - 7) * (x+8) = 0
and the value of x will be 7 or - 8....

x^2y^2-81=(xy-9)(xy+9)


x+x^2=56 x^2+x-56=0 (x+8)(x-7)=0 x=-8 x=7

(x y )² - 81
[ (xy) - 9 ] [ (xy) + 9 ]

x² + x - 56 = 0
(x + 8)(x - 7) = 0
x = 7 is acceptable as a solution.

Hi, i think that the answer to the first question is not actually "an" answer, but two that put together and make the equation. Take one X, one Y from each of the squardeds (?)
and then find the number that times itself would make -81.
it would look like this (X*Y-9) and (X*Y+9), so when you multiply them together, they equal the equation. You didn't specify whether or not you needed to make points for a graoh or not, so i will try to help with that anyway. make one of the equations equal to zero, and solve it. then apply the answer to the other equation, and you'll have your two points.

As for the other question, you also have to make an equation. X+Xsquared= 56 Move the single x to the other side, so its' X2=56-x (cuz its a positive x) then solve both sides. You will get your answer.
I hope that helps you!! P.S. I just took algebra this year, and its tough!

a^2 - b^2 = (a + b)(a - b)

1)
x^2y^2 - 81
= (xy + 9)(xy - 9)

= = = = = = = =

2)
x = the number

x + x^2 = 56
x^2 + x - 56 = 0
x^2 + 8x - 7x - 56 = 0
(x^2 + 8x) - (7x + 56) = 0
x(x + 8) - 7(x + 8) = 0
(x + 8)(x - 7) = 0

x + 8 = 0
x = -8

x - 7 = 0
x = 7

∴ x = -8 , 7

squareroots of 56
the missing number is 7.48

hmm.. I can't figure out the first problem... but to show your work.. see what you did in your head... and if you can't find that out... work backwords.