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avatar+4676 

Help.

 Nov 2, 2017
 #1
avatar+98197 
+2

12x^2 - 11x  = 11

 

12x^2 - 11x - 11 = 0      ⇒  ax^2 + bx  + c  = 0

 

The discriminant is     b^2  - 4ac      

 

(-11) ^2  - 4(12)(-11)  =  121 + 528  >  0

 

So.....when the discriminant > 0....we will have two REAL solutions

 

 

cool cool cool

 Nov 3, 2017
 #2
avatar+98197 
+1

 

 

Mmmmm...on the second one....the discriminant > 0...but I don't see how that decides much of anything

 

I just means that we have real roots

 

Look at this graph :  https://www.desmos.com/calculator/iohx6ebf5u

 

The curve definitely exceeds 141 for a long period of time   !!!!

 

 

cool cool cool

 Nov 3, 2017
 #3
avatar+4676 
0

So the answer is yes?

NotSoSmart  Nov 3, 2017
 #4
avatar+98197 
+1

Yep   .....

 

 

cool cool cool

 Nov 3, 2017
 #5
avatar+4676 
+1

Ok, thanks

NotSoSmart  Nov 3, 2017
 #6
avatar
-1

No, the answer is No.

For T to be greater than 141, you need -0.005x^2 + 0.45x to be greater than 16, so the quadratic that you should be considering is -0.005x^2 + 0.45x - 16 = 0.

If this has two real roots, between these two values of x, T will be greater than 141, otherwise T will never reach 141.

It doesn't, so it doesn't.

(Chris has typed in 0.0005 rather than 0.005).

 

Tiggsy

 Nov 3, 2017

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