ballaholic8
New member
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- Nov 1, 2008
- Messages
- 5
AB is diam of circle w/ center O, P on circle, AP extended
- Thread starter ballaholic8
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
Re: Question
24 is an AWESOME problem!!!
1st thing to notice. If it has TWO Real roots, it must have THREE Real roots. You tell me why.
2nd thing to notice. If it has THREE Real roots, and one of them is a double, it must look like this: \(\displaystyle (x-a)^{2}*(2x-b)\). You tell me why.
Expand that expression and equate coefficients on corresponding powers of x. This leads to:
2 = 2
-9 = -b - 4a
12 = 2ab + 2a^2
-k = -ba^2
Solve this system for a and b and you are just about done.
-9 = -b - 4a
12 = 2ab + 2a^2
24 is an AWESOME problem!!!
1st thing to notice. If it has TWO Real roots, it must have THREE Real roots. You tell me why.
2nd thing to notice. If it has THREE Real roots, and one of them is a double, it must look like this: \(\displaystyle (x-a)^{2}*(2x-b)\). You tell me why.
Expand that expression and equate coefficients on corresponding powers of x. This leads to:
2 = 2
-9 = -b - 4a
12 = 2ab + 2a^2
-k = -ba^2
Solve this system for a and b and you are just about done.
-9 = -b - 4a
12 = 2ab + 2a^2
- Joined
- Apr 12, 2005
- Messages
- 11,339
Re: Question
Oops! I was supposed to use calculus.
If f(x) = 2x^3 -9x^2 + 12x - k
Then f'(x) = 6x^2 - 18x + 12
Further, f'(x) = 0 at x = 1 and x = 2
Solve for k
f(1) = 0 and f(2) = 0
You tell me why that worked AND produced the same, unique solutions.
Rule of Thumb: Algebra was studied well before the calculus. Don't leave your algebra in your desk drawer.
Oops! I was supposed to use calculus.
If f(x) = 2x^3 -9x^2 + 12x - k
Then f'(x) = 6x^2 - 18x + 12
Further, f'(x) = 0 at x = 1 and x = 2
Solve for k
f(1) = 0 and f(2) = 0
You tell me why that worked AND produced the same, unique solutions.
Rule of Thumb: Algebra was studied well before the calculus. Don't leave your algebra in your desk drawer.