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#1**+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

CPhill Nov 3, 2017

#2**+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 !!!!

CPhill Nov 3, 2017

#6**-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

Guest Nov 3, 2017