How do you solve #3x^5 = 6x^4 - 72x^3 =#?

Question

1203 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-22T16:37:08

#(1) :# The Soln., in #RR, x=0;# & in #CC, x=0, x=1+-isqrt23#.

#(2) :# The Reqd. Value, (i.e., "=?" of the problem) in #RR, =0;# and, in #CC, 0, or, -288(-11+-isqrt23)#.

Given that, #3x^5=6x^4-72x^3#

#rArr3x^5-6x^4+72x^3=0#

#rArr 3x^3(x^2-2x+24)=0#

#rArr x^3=0, or, x^2-2x+24=0#

#x^3=0 rArr x=0..........(1)#

#x^2-2x+24=0................(2)#

#rArrx^2-2x=-24#

#rArrx^2-2x+1=-24+1#

#rArr (x-1)^2=-23#, which is not possible in #RR#

However, in #CC, (x-1)^2=-23 rArr x-1=+-isqrt23#

#rArr x=1+-isqrt23#

Hence, in #RR#, the reqd. value (i.e., "=?" of the problem)#=0#

While, in #CC#, the reqd. value#=3x^5=0, if x=0#, or,

if, #x=1+-isqrt23#, the reqd. value (i.e., "=?" of the problem)

#6x^4-72x^3=6x^2(x^2-12x)=6(2x-24)(-24).........[by (2)]#

#=-24*6*2(x-12)=-288(1+-isqrt23-12)=-288(-11+-isqrt23)#