The area of a square, minus its perimeter, would equal 12. Find the length of the square.
So far, I've only dealt with problems that require the use of A = L^2, and in no way included use of the word "perimeter". Help please, and thank you many, many times over.
Word problem dealing with area.?
2008-08-24 6:02 am
回答 (5)
2008-08-24 6:07 am
✔ 最佳答案
If the side length of a square is s, then area is s^2, and its perimeter is 4s. So you're told that s^2 - 4s = 12. You can solve this for s by factoring.s^2 - 4s = 12
s^2 - 4s - 12 = 0
(s - 6)(s + 2) = 0
s = 6, -2
Now a side length of -2 doesn't make any sense. So the side length has to be 6.
2008-08-24 12:34 pm
x = length of the square
x^2 - 4x = 12
x^2 - 4x - 12 = 0
x^2 + 2x - 6x - 12 = 0
(x^2 + 2x) - (6x + 12) = 0
x(x + 2) - 6(x + 2) = 0
(x + 2)(x - 6) = 0
x + 2 = 0
x = -2
x - 6 = 0
x = 6
∴ x = -2 , 6
(the length of the square is 6 units.)
x^2 - 4x = 12
x^2 - 4x - 12 = 0
x^2 + 2x - 6x - 12 = 0
(x^2 + 2x) - (6x + 12) = 0
x(x + 2) - 6(x + 2) = 0
(x + 2)(x - 6) = 0
x + 2 = 0
x = -2
x - 6 = 0
x = 6
∴ x = -2 , 6
(the length of the square is 6 units.)
2008-08-24 6:07 am
perimeter is the length of its boundary.
Here sq is in sq units and perimeter in units. How can one subtract one from other?
You can solve x^2 - 4x = 12 but the wording certainly is incorrect.
Here sq is in sq units and perimeter in units. How can one subtract one from other?
You can solve x^2 - 4x = 12 but the wording certainly is incorrect.
2008-08-24 6:06 am
x^2 - 4x = 12
x^2 -4x -12 = 0
(x-6)(x+2) = 0
x =6 (since -2 is not possible)
x^2 -4x -12 = 0
(x-6)(x+2) = 0
x =6 (since -2 is not possible)
2008-08-24 6:06 am
im confuzled
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