SOLUTION: Please help solv. A digital photgraph is 30 pixel longer than it is wide and has a total area of 4000 pixels. Find the dimensions of this photograph.

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Question 555266: Please help solv. A digital photgraph is 30 pixel longer than it is wide and has a total area of 4000 pixels. Find the dimensions of this photograph.
Found 2 solutions by jim_thompson5910, nerdybill:
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
x(x+30)=4000

x^2+30x=4000

x^2+30x-4000=0

Now use the quadratic formula to solve:

x = (-b+-sqrt(b^2-4ac))/(2a)

x = (-(30)+-sqrt((30)^2-4(1)(-4000)))/(2(1))

x = (-30+-sqrt(900-(-16000)))/(2)

x = (-30+-sqrt(16900))/2

x = (-30+sqrt(16900))/2 or x = (-30-sqrt(16900))/2

x = (-30+130)/2 or x = (-30-130)/2

x = 100/2 or x = -160/2

x = 50 or x = -80

Ignore the negative solution. So the dimensions are 50 by 80 pixels.

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Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
Please help solv. A digital photgraph is 30 pixel longer than it is wide and has a total area of 4000 pixels. Find the dimensions of this photograph.
Let x = width (in pixels)
then
x+30 = length
.
x(x+30) = 4000
x^2+30x = 4000
x^2+30x-4000 = 0
(x+80)(x-50) = 0
x = {-80, 50}
throw out the negative solution leaving:
x = 50 pixels (width)
.
Length:
x+30 = 50+30 = 80 pixels
.
Dimensions:
50 by 80 pixels