SOLUTION: A rectangular frame has a length of (x+3) units and width (x-4) if the area of the frame is 4 sq units what is the length of the frame?

Algebra ->  Surface-area -> SOLUTION: A rectangular frame has a length of (x+3) units and width (x-4) if the area of the frame is 4 sq units what is the length of the frame?      Log On


   

Question 921135: A rectangular frame has a length of (x+3) units and width (x-4) if the area of the frame is 4 sq units what is the length of the frame?
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
a length of L=%28x%2B3%29 units and width W=%28x-4%29 if the area of the frame is A=4 sq units
A=L%2AW
4=%28x%2B3%29%2A%28x-4%29
4=x%5E2-4x%2B3x-12
4=x%5E2-x-12
0=x%5E2-x-12-4
0=x%5E2-x-16........use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2-4%2A1%2A%28-16%29+%29%29%2F%282%2A1%29+
x+=+%281+%2B-+sqrt%28+1%2B64+%29%29%2F2+

x+=+%281+%2B-+sqrt%2865+%29%29%2F2+
x+=+%281+%2B-+8.06225774829855%29%2F2+
x+=+%281+%2B-+8.06%29%2F2+
solutions:
x+=+%281+%2B+8.06%29%2F2+
x+=+9.06%2F2+
x+=+4.53
or
x+=+%281-+8.06%29%2F2+
x+=-7.06%2F2+
x+=+-3.53..........we don't need this solution because the length and width cannot be negative number
L=%28x%2B3%29 units => L=%284.53%2B3%29 units=> L=7.53 units
and
W=%28x-4%29 units=>W=%284.53-4%29 units=>W=0.53 units
check the area
4=7.53%2A0.53
4=3.9909
4=4