No approximations please.
Exact roots only.
Radicals allowed.
2006-12-20 03:41:19 · 3 answers · asked by oscarD 3 in Science & Mathematics ➔ Mathematics
ghakh i think you mean -2.
2006-12-20 03:55:47 · update #1
how about the rest?
2006-12-20 03:56:27 · update #2
noticing that x+2 is a root and the leading term is 16x^5 may give an hint...
2006-12-20 04:00:41 · update #3
hey thanks Joshua Z but i know the answer- this one is for 10 points to the first to solve..
2006-12-20 04:08:13 · update #4
hey pascal, this one was with you in mind too (one of your prior questions)!
2006-12-20 04:09:15 · update #5
the 10 i've gotta award to a derivation (if submitted) tho...
2006-12-20 04:10:16 · update #6
what is f(x-2)?
2006-12-20 04:27:03 · update #7
pretty cool pascal..
who remembers the expansion of cos^5(x) ?
2006-12-20 07:28:50 · update #8
x = -√(5-√5)/(2√2)-2
x = √(5-√5)/(2√2)-2
x = -√(√5+5)/(2√2)-2
x = √(√5+5)/(2√2)-2
x = -2
Edit: since you asked so politely -
First, we convert the quintic into a depressed quintic using the substitution x=t-2:
16 (t-2)^5 + 160 (t-2)^4 + 620 (t-2)³ + 1160 (t-2)² + 1045 (t-2) + 362 = 0
Exapnding this:
16 (t^5 - 10t^4 + 40t³ - 80t² + 80t - 32) + 160 (t^4 - 8t³ + 24t² - 32t + 16) + 620 (t³ - 6t² + 12t - 8) + 1160 (t^2 - 4t + 4) + 1045 (t-2) + 362 = 0
16t^5 - 160t^4 + 640t³ - 1280t² + 1280t - 512 + 160t^4 - 1280t³ + 3840t² - 5120t + 2560 + 620t³ - 3720t² + 7440t - 4960 + 1160t² - 4640t + 4640 + 1045t - 2090 + 362 = 0
Combing like terms:
16t^5 + 20t^3 + 5t = 0 (phew, that's much simpler to work with)
This factors as t(16t^4 + 20t^2 + 5) = 0. The latter equation is a biquadratic, so we make the substitution z=t²:
16z² + 20z + 5 = 0
By the quadratic formula:
z= (-20 ± √(400 - 320))/32 = (-5 ± √5)/8
So t=±√(-5±√5)/√8 or t=0
And x=±√(-5±√5)/√8 - 2 or x=-2
Trying every possible sign pattern on the first solution yields the first four solutions I posted at the beginning.
2006-12-20 04:07:34 · answer #1 · answered by Pascal 7 · 1⤊ 0⤋
Well, x = -2 works, and the polynomial then factors into
(x+2)(16x^4 + 128x^3 + 364x^2 + 432x + 181) = 0,
and since there's a quartic formula, you can proceed from there. One of the solutions is very close to -sqrt(2) but that's not exact unfortunately.
For the quartic formula written out in all its glory, visit
http://planetmath.org/encyclopedia/QuarticFormula.html
and if you'd like to work through all the substitutions yourself along the way, you might like
http://en.wikipedia.org/wiki/Quartic_equation
I think the best explanation might be the one from Ask Dr Math at
http://mathforum.org/dr.math/faq/faq.cubic.equations.html
which covers both cubics and quartics.
Enjoy!
2006-12-20 12:05:00 · answer #2 · answered by Joshua Z 1 · 1⤊ 0⤋
use synthetic division ot solve this, 2 is one of the roots.
2006-12-20 11:50:08 · answer #3 · answered by ghakh 3 · 0⤊ 0⤋