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Help please <3

 Apr 26, 2018

If the question is asking you to make W the subject, this is my method

W= x

length = 2x + 8

as it is double W + 8

and we know that width times the length is 72 

therefore 72/lenth must be W

W= 72/2x+8

 Apr 26, 2018

Let w = width of shoebox

Let l = length of the shoebox


Since the length is 8 centimeteres more than twice the width, we can conclude that l = 2w+8. 


The area of a rectangle is given, so we can solve for the now one-variable equation. 


\(w(2w+8)=72\)Expand completely.
\(2w^2+8w-72=0\)Factor out the GCF of every term, which is 2 in this case.
\(2(w^2+4w-36)=0\)Unfortunately, this trinomial is not factorable. Let's use the quadratic formula instead. 
\(a=1,b=4,c=-36;\\ x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\)Now, this is a matter of simplification.
\(x_{1,2} = {-4 \pm \sqrt{4^2-4(1)(-36)} \over 2(1)}\) 
\(x_{1,2} = {-4 \pm \sqrt{160} \over 2}\)It is possible to simplify the radical. 
\(x_{1,2} = {\frac{-4\pm4\sqrt{10}}{2}}\)Simiplify the fraction as every term has a common factor of 2.
\(x_{1,2} = -2\pm2\sqrt{10}\)Reject the negative answer since a width can never be negative in the context of geometry.
\(x=-2+2\sqrt{10}\approx 4.32 \text{cm}\) 


When I read the question more closely, I realized that all this solving was completely superfluous. You can still use the equation from the beginning, though! It represents the situation perfectly.

 Apr 26, 2018
edited by TheXSquaredFactor  Apr 26, 2018

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