Question

how many real solutions does the equation 0.2x^5-2x^3+1.8x+k=0. have when k=0. a. how many real solutions when it k=1. b.is there a value for k for which the equation has just one real solution? b. is there a value for k for which the equation has no real solution?

Answer 1

When k =0:
\[0.2x^{5}-2x^{3}+1.8x=0\]\[x(0.2x^{4} - 2x^{2}+1.8)=0\]\[0.2x^{4}-2x^{2}+1.8=0\]Suppose y = x^2. Now,
\[0.2y^{2}-2y+1.8=0\]
Comparing it with ay^2+by+c=0:
a=0.2
b=-2
c=1.8
\[y = \frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]\[y = \frac{ 2\pm \sqrt{4-(4\times0.2\times1.8)} }{ 2\times0.2 }\]\[y = \frac{ 2\pm \sqrt{2.56} }{ 0.4 }\]Now using (+ve) and (-ve) sign find out the soultion for y and then put the results here:\[y = x^{2}\]You will get two values of x. Follow the same process for next one.

Answer 2

When k=1 you can't aggregate the x... @curiousshubham

Answer 3

:(

Answer 4

my head is burning...........

Answer 5

y

Answer 6

Today I read a lot.... and em feeling sleepy..

Answer 7

good night

Answer 8

you too.......

Answer 9

n u?

Answer 10

I feel like I'm trying to create a new math formula XD

Answer 11

@decripter37 whats that?

Answer 12

I following this path: ax^5+bx^3+cx=k => x(ax^4+bx^2+c)=k => x=k/y, ax^4+bx^2+c=y and I'm trying to find y :)

Answer 13

keep it up...

Answer 14

You can also draw the function, it's easy you just have to find max, min with derivatives, and after see how many times the function pass trough the 1

Answer 15

ok thnks