SOLUTION: I have a world problem, not sure if I solved it correctly. Could I have some help, thank you. A picture frame is 14cm by 18cm. There are 192cm2 of the picture showing. Find the wid

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: I have a world problem, not sure if I solved it correctly. Could I have some help, thank you. A picture frame is 14cm by 18cm. There are 192cm2 of the picture showing. Find the wid      Log On


   



Question 190555: I have a world problem, not sure if I solved it correctly. Could I have some help, thank you. A picture frame is 14cm by 18cm. There are 192cm2 of the picture showing. Find the width of the frame. I have 2cm. Is this correct. Thank you.
Found 3 solutions by nerdybill, nene189, stanbon:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = width of frame
then
(14-2x)(18-2x) = 192
(14-2x)(18-2x) = 192
252 - 28x -36x + 4x^2 = 192
252 - 64x + 4x^2 = 192
4x^2 - 64x + 252 = 192
4x^2 - 64x + 60 = 0
x^2 - 16x + 15 = 0
(x-15)(x-1) = 0
x = {1,15}
.
We can toss out the 15 leaving us with:
x = 1 cm

Answer by nene189(2) About Me  (Show Source):
You can put this solution on YOUR website!
14 * 18= 252cm2 so 2cm is not the answer

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A picture frame is 14cm by 18cm. There are 192cm2 of the picture showing. Find the width of the frame. I have 2cm.
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Draw a rectangle inside a rectangle.
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Let the uniform frame width be "x".
(18-2x)(14-2x) = 192
252-28x-36x+4x^2 = 192
4x^2-64x+60 = 0
x^2-16x+15) = 0
(x-15)(x-1)=0
Realistic solution:
x = 1 cm
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Cheers,
Stan H.