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Xelga [282]
4 years ago
15

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

Mathematics
1 answer:
NARA [144]4 years ago
5 0
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
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