OK so i was given this problem..
p(x) = x^{4} + 4x^{2}
can i use the quadratic formula for this problem? even though it doesn't have a third term?
Last edited by abel muroi; 03-10-2015 at 09:41 PM.
You are not asking a question. Are you trying to find p(5). If so, then you do not need to use the quadratic equation. If you want to find the y-intercept then you let x=0 and do not use the quadratic formula. If you want to know what happens to p(x) when x is very large you do not use the quadratic formula.
In fact you never have to use the quadratic formula.
Also most people would only think about using the quadratic formula when the function is quadratic not cubic.
You seem to have the idea that given a particular function that there is just one thing to do to it so there is no need to ask a question because the question is obvious. Move away from that thinking.
Good luck.
A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin
Abel, all you did was change the function! You still are not asking a question. For the record p(1) = 5 and I did not use the quadratic formula.
A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin
x^4 + 4x^2 = 0
x^2(x^2 + 4) = 0
so:
x^2 = 0 or x^2 + 4 = 0
I'm just an imagination of your figment !
I think you mean "the quadratic formula". You can use the quadratic formula for any quadratic equation to find the solutions. The solutions are the x-coordinates of the intersection of the curve with the x-axis. When the curve does not cross the x-axis you get a minus sign inside the square root and you cannot take a minus sign of a square.
Last edited by stapel; 03-13-2015 at 10:48 AM. Reason: Fixing carriage returns.
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