Explain how to find the zeros of the function f(x) = 2x3 – 9x + 3. Help, please:)
f(x)=2x^3-9x+3
Let x = (a+b). Then x^3 = (a+b)^3 = a^3 + 3 a b (a + b) + b^3 = a^3 + 3 a b x+ b^3, or x^3 - 3 a b x - (a^3+b^3) = 0 Solve 3 a b = 9, and - (a^3+b^3) = 3 Then the answer is x = a + b x1= 3^(2/3)/(1/2 (-1 + i sqrt(11)))^(1/3) + (3/2 (-1 + i sqrt(11)))^(1/3) x2 = -(((3/2)^(2/3) (1 +i sqrt(3)))/(-1 +i sqrt(11)^(1/3)) - 1/2 (1 -i sqrt(3)) (3/2 (-1 +i sqrt(11)))^(1/3) x3 = -(((3/2)^(2/3) (1 -i sqrt(3)))/(-1 +i sqrt(11))^(1/3)) - 1/2 (1 +i sqrt(3)) (3/2 (-1 + i sqrt(11)))^(1/3)
Does that help, a little? :3
It's a bit confusing but I'll find a way. Thanks you:)
If anyone else has different methods of solving something like this please share:)
@jackoo two questions: 1. Are you sure you copied the problem correctly (I tried a few variations of it, none have "nice" roots. 2. What sort of answer are you looking for? Deriving the cubic equation? Numerical approximation? Graphing?
write it out in factored form, then mulitply out. say you want the zeros to be -2, 3 and 1/2 the write out (x+2)(x−3)(2x−1)
that obviously has zeros at -2, 3 and 1/2. multiply out to put it in standard form. i get 2x^3−3x^2−11x+6 but you can make up your own
just pick the zeros, write in factored form, and then multiply out. much easier than last problem, yes.
Yes, I wrote it incorrectly but I corrected the equation. f(x)=2x^3-9x+3
1) write in factored form (don't show anyone) 2) multiply it out 3) factor again, you will get what you started with 4) now you know the zeros
Thank you @dumbsearch2 and @mathteacher1729 do you have a different method?
(BTW if it helps, I would greatly appreciate if you could mark my answer as correct. ^_^)
Will do:)
:) Oh, and you'll find all the answers you need here: https://www.google.com/search?q=f(x)+%3D+2x3+%E2%80%93+9x+%2B+3&oq=f(x)+%3D+2x3+%E2%80%93+9x+%2B+3&aqs=chrome.0.57j62l3j60.305&sourceid=chrome&ie=UTF-8#hl=en&safe=off&tbo=d&sclient=psy-ab&q=f(x)+%3D+2x3+%E2%80%93+9x+%2B+3+site:openstudy.com&oq=f(x)+%3D+2x3+%E2%80%93+9x+%2B+3+site:openstudy.com&gs_l=serp.3...4270.6327.0.6503.19.19.0.0.0.0.211.2828.0j16j2.18.0.les%3B..0.0...1c.1.3.psy-ab.a95BNdOitko&pbx=1&bav=on.2,or.r_gc.r_pw.r_qf.&bvm=bv.42553238,d.aWc&fp=9d6268e7630cf0ae&biw=1280&bih=624 You'd be suprised about how a simple Google search helps you find everything. :)
A lot of questions here have been asked previously.
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