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 #2
avatar+4476 
+1

\(\frac{1}{x(x-4)}+\frac{5-2x}{x^2-3x-4}=\frac{5}{x(x+1)}\)

 

Factor the  x2 - 3x - 4 . What two numbers add to  -3  and multiply to  -4 ?       +1  and  -4  .

 

\(\frac{1}{x(x-4)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}\)

                                                                      Multiply the first fraction by  \(\frac{x+1}{x+1}\) .

\(\frac{1(x+1)}{x(x-4)(x+1)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}\)

                                                                      Multiply the middle fraction by  \(\frac{x}{x}\) .

\(\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5}{x(x+1)}\)

                                                                      Multiply the last fraction by  \(\frac{x-4}{x-4}\) .

\(\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5(x-4)}{x(x+1)(x-4)}\)

 

Now we have a common denominator. Let's multiply both sides by this denominator...but first we must say that   x ≠ 0 ,  x ≠ 4 ,  and  x ≠ -1 , because these values cause a zero in the denominator of the original equation.

 

1(x + 1) + (5 - 2x)x  =  5(x - 4)          When    x ≠ 0 ,  x ≠ 4 ,  and  x ≠ -1  .

                                                             Multiply out the parenthesees and simplify.

x + 1 + 5x - 2x2  =  5x - 20

 

-2x2 + x + 21  =  0

                                                             Solve for  x  with the quadratic formula.

\(x = {-1 \pm \sqrt{1^2-4(-2)(21)} \over 2(-2)} \\~\\ x = {-1 \pm 13\over -4} \\~\\ x=-\frac{12}{4}=-3\qquad\text{ or }\qquad x=\frac{-14}{-4}=\frac{7}{2}\)

hectictar 12-sep-2017
 #1
avatar+4476 
+1

Here's  a,  b,  and  c .

 

a)

volume of box  =  (length )(  width )(height)

          V            =  (25 - 2x)(20 - 2x)(x)

          V            =  (500 - 90x + 4x2)(x)

          V            =  500x - 90x2 + 4x3

          V            =  4x3 - 90x2 + 500x

 

Since  x  is folded up from a side that is 20 inches long, x can't be larger than 10 inches. If x were 10 inches, it would fold 20 in half, and the width would be 0 .  So... the domain is all real  x | 0 ≤ x ≤ 10 .

 

 

b)

To find the side length of the squares (x) that yield a volume of 300 in3 ,

plug in  300  for  V and solve for  x .

 

300  =  4x3 - 90x2 + 500x

 

Here I used a graph to find the approximate solutions.

x ≈ 0.68 inches  and  x ≈ 7.93 inches  are the only values in the domain.

 

So...the size of the corner squares can be....

≈ 0.68 by 0.68    or     ≈ 7.93 by 7.93

 

 

c)

We can look at the graph again to see that the x values that cause a volume bigger than 300 are those between 0.68 and 7.93 , and those greater than 13.89 .

 

Since those greater than 13.89 are outside the domain, the x values that cause the volume to be bigger than 300 are......          0.68 < x < 7.93 .

 

So....for instance, the size of the corner square could be  0.5"  by  0.5",  or  1"  by  1", or  3"  by  3".

hectictar 11-sep-2017