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

Question

Mathematics

Lennon Whitney

15 February, 01:31

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

+4

Answers (1)

Know the Answer?

Sasha Richmond · Answer 1

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