Question 177744
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The new computer can do the job in <i>x</i> hours. So the new computer can do *[tex \Large \frac{1}{x}]th of the job in one hour.


The old computer can do the job in <i>x </i>+ 1 hours.  That means that the old computer can do *[tex \Large \frac{1}{x + 1}]th of the job in one hour.



Working together, the two computers can do *[tex \Large (\frac{1}{x})+(\frac{1}{x + 1})]th of the job in one hour.


Perform the sum:


*[tex \Large (\frac{1}{x})+(\frac{1}{x + 1})=\frac{(x + 1)+ x}{x^2 + x}=\frac{2x + 1}{x^2 + x}]


Take this expression for the portion of the job that can be completed by both computers working together for one hour and take the reciprocal to find an expression of how long it would take both computers working together to perform the entire job:


*[tex \Large \frac{1}{\frac{2x + 1}{x^2 + x}}=\frac {x^2 + x}{2x + 1}]


We are given that this total time is 3 hours, so set the above expression equal to 3:


*[tex \Large \frac {x^2 + x}{2x + 1}=3]


Multiply both sides by *[tex \Large 2x + 1]:


*[tex \Large x^2 + x = 6x + 3]


Collect all terms on the left and combine like terms:


*[tex \Large x^2 - 5x - 3 = 0]


You need to either complete the square or use the quadratic formula to solve for x which represents the time it takes for the new computer to do the job by itself.  Note, you will get two solutions for x, but one of them will be less than zero.  This absurd result is an extraneous root caused by squaring the variable in the process of solving the problem.  Exclude the negative root and the other root will be the answer.


Since you don't mention any required round-off precision, I would leave the answer in radical form, that being the only way to represent the exact answer.










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