Raymond B. answered • 04/28/24
Math, microeconomics or criminal justice
2 ways to view this problem
1st is one way, but almost like going the wrong way on a one way street
2nd way is likely what you want, skip on past the jump car to get to #2 though
5th roots of
x^4(x-4)^2
= (x^4(x-4)^2))^(.2) = y = 0
use a graphing calculator
and see where the x intercepts are, if any
if none, they're imaginary
the graph is like a small hill inside a steep valley
only y intercept is the origin (0,0)
local max is atop the small hill. local&global mins are (0,0) and (4,0)
no global maximum
(0,0) and (4,0) are real x intercepts = real roots, x=0, 4
other roots are necessarily either repetitive or imaginary
x=0 looks repetitive, 0,0,0,0
a 5th and 6th root is x= 4, repetitive: 4,4
(x^4)(x^2-8x+16)
= x^6 -8x^5 +16x^4
by Decartes' rule of signs, 2 sign changes, 2 positive real roots
(-x)^6- 8(-x)^5 +16(-x)^4
= x^6 +8x^5+16x^4 has 0 sign changes, 0 negative real roots
jump start the stalled car, especially in cold weather, or call Triple A
#2 solution, a work in progress, check back later for more progress
just 5 roots in this solution. square roots have 2 roots, cube roots have 3 roots
"quad roots" 4, "quin roots" 5, although some may be real, repetitive or imaginary roots, some negative, some positive, with an "x" in the mix, you get to get all of them all at once, hard trick to do, but math is magic
(x^4)(x-4)^2)^(1/5)
= x^(4/5)(x-4)^(2/5)
=x^.8(x-4)^.4
= (back in a few minutes, this needs more work, plus there's 5 roots, so you can't stop at 1. slow down at 3 and especially at 4, as the light is yellow ready to turn red. where are the other tutors on this? why the no shows?)
