Question

How many real solutions does the equation 0.2x^5-2x^3+1.8x+k=0 have when k = 0?

Answer 1

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

Explanation:

Edg 2020

Answer 2

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