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what is the answer to the problem    x(7-x)>8

please give me the answer I have tried to fing it for 3 days and everyone says no solution but when I go back to the question in the Pearson textbook.. even photomath says that there is no answer

 

 

 

Thanks and as always, Have a WONDERFUL afternoon,

                                                                                         a math student that is having a lot of trouble  

 Oct 21, 2015

Best Answer 

 #4
avatar+130560 
+5

 x(7-x)>8   simplify the left side

 

7x - x^2 > 8     subtract 8 from both sides

 

7x - x^2 - 8 > 0     multiply through by -1 and remember to reverse the inequality sign

 

-7x + x^2 + 8 < 0   rearrange

 

x^2 - 7x + 8 < 0      

 

Here's a graph of this parabola :  https://www.desmos.com/calculator/jcwal2dhty

 

Note that the interval where the function is < 0  is about (1.438, 5.562)....and this is the solution

 

And, as "Guest" has said, the exact interval is : 1/2 (7-sqrt(17))<x<1/2 (7+sqrt(17))

 

 

cool cool cool

 Oct 22, 2015
edited by CPhill  Oct 22, 2015
 #1
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0

i cant find the answer sorry

 Oct 21, 2015
 #2
avatar
+5

  x(7-x)>8

 

This is the expanded form:7x - x^2>8, and x=2, 3, 4, 5

This is the solution:1/2 (7-sqrt(17))<x<1/2 (7+sqrt(17))

 Oct 21, 2015
 #3
avatar
0

there really isn't a answer unless u figure out x

 Oct 21, 2015
 #4
avatar+130560 
+5
Best Answer

 x(7-x)>8   simplify the left side

 

7x - x^2 > 8     subtract 8 from both sides

 

7x - x^2 - 8 > 0     multiply through by -1 and remember to reverse the inequality sign

 

-7x + x^2 + 8 < 0   rearrange

 

x^2 - 7x + 8 < 0      

 

Here's a graph of this parabola :  https://www.desmos.com/calculator/jcwal2dhty

 

Note that the interval where the function is < 0  is about (1.438, 5.562)....and this is the solution

 

And, as "Guest" has said, the exact interval is : 1/2 (7-sqrt(17))<x<1/2 (7+sqrt(17))

 

 

cool cool cool

CPhill Oct 22, 2015
edited by CPhill  Oct 22, 2015

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