Respuesta :
when k = we have:-
0.2x^5 - 2x^3 + 1.8x = 0
so x = 0 is one solution
taking 0.2x out we have:-
0.2x( x^4 - 10x^2 + 9x) = 0
factoring::-
(x^2 - 9)(x^2 - 1) = 0
this gives x = +/= 3 and x = +/- 1
so there are 5 real solutions
0.2x^5 - 2x^3 + 1.8x = 0
so x = 0 is one solution
taking 0.2x out we have:-
0.2x( x^4 - 10x^2 + 9x) = 0
factoring::-
(x^2 - 9)(x^2 - 1) = 0
this gives x = +/= 3 and x = +/- 1
so there are 5 real solutions
Answer:
It is 5
When Y=1 I got 0
Is there a value of k for which the equation has just one real solution?
I put no
Then for the last one I put Yes
Step-by-step explanation:
Edg 2020
