Question

help !

Answer 1

Yes?

Answer 2

consider the both equation \[x ^{5} = 16x ^{3}\]

dividing both sies by x3 gives \[\frac{ x^5 }{ x^3} = \frac{ 16x^3 }{ x^3 }\]

i.e
and hence
however, as well as these two solutions of x = 4 and x = -4 there is another value for x that satisfies the original equation. what is this 3rd solution tot he original equation?

Answer 3

x = 0? :P

Answer 4

i.e \[x^2 = 16 \]
and hence \[x = \pm4\]

Answer 5

instead of dividing each side of the original equation by x^3 to solve it, instead subtract \[16x^3\] from each side and then factories and solve

Answer 6

see if u just put x = 0 , that wud satisfy the original equation , wont it ? :S

Answer 7

whatcha think?

Answer 8

so three solutions wud be x = 4,-4 and 0

Answer 9

@RANE

Answer 10

if a queston is in that format don't do that by that approach, that can give you extraneous roots..... use this approach x^5-16x^3=0 then find the roots

Answer 11

i was just wondering if we can put x =0 that will satisfy the original equation i.e.
0^5 = 16 0^2
0 = 0
satisfied
dont u think so? @divu.mkr

Answer 12

but we are dividing x^2 both sides and if we put x=0 that will become invalid..and solution for every root is same

Answer 13

x^5 = 16x^3
x^5 - 16x^3 = 0

Answer 14

then factor

Answer 15

thank you all of you