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Can someone tell me what I'm doing wrong here? Long day.
f(x) = 2x² + px + q. Given that f(-3) = 0, and f(4) = 2, find the value of p and q.
f(x) = 2x² + px + q.
f(-3) = 2(-3)² - 3p + q = 0
= 18 - 3p + q = 0
∴ q = 3p - 18
f(4) = 2(4)² + 4p + q - 2 = 0
= 30 + 4p + q = 0
substituting the first into the second:
30 + 4p + (3p - 18) = 0
30 + 4p + 3p - 18 = 0
7p = - 12
p = -12/7
I know I haven't given an answer for q... only because my answer for p is wrong. What have I done?
Thanks
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Edited by Pi Man: Ignore this post (#2) . I made the mistake, not Daniel.
f(4) = 2(4)² + 4p + q - 2 = 0
= 30 + 4p + q = 0
f(4) = 2(4)² + 4p + q - 2 = 0
= 32 + 4p + q = 0 32, not 30
q = -4p -32
-4p-32 = 3p - 18
-14 = 7p
p = -2
Last edited by pi man (2007-10-22 16:27:28)
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No, Daniel is right. 2(4²) -2 = 30, like he says.
In fact, everything is right, including the value of p. Be more confident!
Why did the vector cross the road?
It wanted to be normal.
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Whoops. Sorry about that Daniel. And thanks, Mathsyperson. I was a little too hasty. If I solved it all the way through and then double-checked, I would have seen I was the one who made the mistake
I'll do now what I should have done before. Continuing with what you already figured out....
p = -12/7 and q = 3p - 18
q = 3(-12/7) - 18
q = -36/7 - 126/7 = -162/7
Plug p and q back into your original question to double-check.
f(x) = 2x² - 12x/7 -162/7
f(-3) = 2(9) + 36/7 - 162/7
= 18 - 126/7
= 18 - 18 = 0
f(4) = 2(16) - 48/7 - 162/7
= 32 - 210/7
= 32 - 30 = 2
Much better!
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Stupid textbook! I waste so much time trying to find where I've gone wrong when actually I haven't!
Thanks!
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What was the book?
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Heinemann AS modular books. The number of errors they make in the answers is ridiculous.
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