SOLUTION: This relates to unknown lenght & width of a rectangle with a known area: How do I solve the equation? (How do I work out what formula to use?). Area = 352 sq.cm 1 side = (X x

Algebra ->  Rectangles -> SOLUTION: This relates to unknown lenght & width of a rectangle with a known area: How do I solve the equation? (How do I work out what formula to use?). Area = 352 sq.cm 1 side = (X x       Log On


   

Question 74628: This relates to unknown lenght & width of a rectangle with a known area:
How do I solve the equation? (How do I work out what formula to use?).
Area = 352 sq.cm
1 side = (X x 2)
other side = (X + 9)
Iwant to show workings - rather than guessing at what X represents.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

YOUR PROBLEM IS NOT CLEAR.
What does%28Xx2%29represent? Is it X%5E2 or is it 2X? I WILL ASSUME 2X
The area(A) of a rectangle equals Length(L) times Width(W)
Either 2X is the length and (X+9) is the width or visa versa--it makes no difference which we choose. So our equation to solve is:
A=L*W
352=2x%28x%2B9%29 get rid of parens
352=2x%5E2%2B18x divide both sides by 2
176=x%5E2%2B9x subtract 176 from both sides
x%5E2%2B9x-176=176-176
x%5E2%2B9x-176=0 quadratic in standard form
A=1
B=9
C=-176
We'll solve using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-9+%2B-+sqrt%28+9%5E2-4%2A1%2A%28-176%29+%29%29%2F%282%2A1%29+
x+=+%28-9+%2B-+sqrt%28+81%2B704%29%29%2F%282%29
x+=+%28-9+%2B-+sqrt%28785%29%29%2F%282%29
x+=+%28-9+%2B-+28.018%29%2F%282%29
x+=+%28-9+%2B+28.018%29%2F%282%29=+9.509 cm
2x=2%2A9.509=19.018 cm------------------------------length
%28x%2B9%29=9.509%2B9=18.509 cm-----------------------------width
We will discount the negative solution for x since lengths and widths are not negative numbers.
CK
A=L*W
352 sq cm=(19.018)(18.509)
352=~352

Hope this helps----ptaylor