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grace14 · Mar 30, 2009, 09:54 AM

Im in the 8th grade and I didn't understand how to do this word problem. Could someone help me?
I haven't started coefficiants or anything like that.. or whatever they are called.

The length of a photograph is 1cm less then twice the width. The area is 45cm squared. Find the dimensions of the photograph.

I think the equation is x(2x-1) = 45. Im not sure what to do from there.

Perito · Mar 30, 2009, 01:07 PM

Originally Posted by grace14

Im in the 8th grade and I didnt understand how to do this word problem. Could someone help me?
I havnt started coefficiants or anything like that..or whatever they are called.

The length of a photograph is 1cm less then twice the width. The area is 45cm squared. Find the dimensions of the photograph.

I think the equation is x(2x-1) = 45. Im not sure what to do from there.

W = width of photograph
L = length of photograph
A = area of photograph

L = 2W - 1 (the length of a photograph is 1 cm less than (not then) twice the width)
A = 45 = L * W (the area is 45 cm^2).

You can substitute the value of L from the first equation (2W-1) into the second equation where you see L. You then have only one equation and one variable (W):

45 = (2W-1) * W

You can multiply this out

$45 = 2W^2 - W$

You can then move the 45 over to the right

$0 = 2W^2 - W - 45$

Factor that and set each factor to zero to find the two possible roots. You can also use the quadratic formula. You then have to decide whether both roots make sense (whether or not there are actually two possible solutions) or whether one of the roots is physically impossible (a negative length) and therefore has no real significance and can be ignored. Once you've found the width, enter it into one of the other equations to calculate the length. Then, to make sure you did everything correctly, use the remaining equation as a check.