- #1
garytse86
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Can someone help me please with inequalities, I have been attempting this question quite a few times but I still can't get the same answer as the textbook.
What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?
Here is my best attempt:
p(x^2+x) < 2x^2 + 6x +1
px^2 + 2p - 2x^2 - 6x -1 < 0
(p-2)x^2 - 6x + 2p -1 < 0
For real roots b^2 - 4ac >= 0
so 6^2 - 4 * (p-2) * (2p-1) >= 0
36 - (8p^2 - 20p + 8) >= 0
-8p^2 + 20p + 28 >= 0
root 1: (-20 + (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 1: -20/-16 + (1296^1/2)/16
root 1: 400/256 + 1296/256
= 1696/256
root 2: (-20 - (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 2: -20/-16 - (1296^1/2)/16
root 2: 400/256 - 1296/256
= -896/256
That's as far as I have got...
the answer in the textbook is p < -1
Can you please help... Thanks a lot.
Gary.
What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?
Here is my best attempt:
p(x^2+x) < 2x^2 + 6x +1
px^2 + 2p - 2x^2 - 6x -1 < 0
(p-2)x^2 - 6x + 2p -1 < 0
For real roots b^2 - 4ac >= 0
so 6^2 - 4 * (p-2) * (2p-1) >= 0
36 - (8p^2 - 20p + 8) >= 0
-8p^2 + 20p + 28 >= 0
root 1: (-20 + (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 1: -20/-16 + (1296^1/2)/16
root 1: 400/256 + 1296/256
= 1696/256
root 2: (-20 - (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
root 2: -20/-16 - (1296^1/2)/16
root 2: 400/256 - 1296/256
= -896/256
That's as far as I have got...
the answer in the textbook is p < -1
Can you please help... Thanks a lot.
Gary.