Question

The number of negative roots of the equation 2x^5 + 7x^3 + 6x^2 + 21=0 is

Answer 1

@mahmit2012 @timo86m @experimentX @stormfire1

Answer 2

factor

(x^3+3)*(2*x^2+7)

Answer 3

thus

(x^3+3)=0
(2*x^2+7)=0

Answer 4

negative roots?

Answer 5

I am getting to it :P

Answer 6

ok

Answer 7

just one

Answer 8

(x^3+3)*(2*x^2+7) the second part has no real values

Answer 9

first (x^3+3)=0

x^3=-3

x=cuperoot(-3)
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second

(2*x^2+7)=0
2x^2=-7
x^2=-7/2
x=sqrt(-7/2)

Answer 10

it has 2 negative roots

Answer 11

they always come in pairs

Answer 12

Raph dont they always come in pairs?

so it is 2?

Answer 13

[x = (1/2*I)*sqrt(14)], [x = -(1/2*I)*sqrt(14)]

Answer 14

sqrt(-7/2)is complex

Answer 15

The answer is 1

Answer 16

they have one complex root

Answer 17

this equation has 2 complex roots. 2 imaginary numbers

Answer 18

one negative root and two complex

Answer 19

here is the whole sol set

[x = -3^(1/3)], [x = (1/2)*3^(1/3)-(1/2*I)*3^(5/6)], [x = (1/2)*3^(1/3)+(1/2*I)*3^(5/6)], [x = (1/2*I)*sqrt(14)], [x = -(1/2*I)*sqrt(14)]

Answer 20

complex roots in real polynomials always come in pairs

Answer 21

aaa ok :)

Answer 22

by negative root he means a solution the the equation where x will be negative?

like -x?

Answer 23

because the sum of it must be a real so they are always conjugates

Answer 24

(x^3+3)*(2*x^2+7)
\[x=-3^{1/3}\]
and two other complex

Answer 25

One negative real root.\[x=-\sqrt[3]{3}=-1.44225 \]A plot is attached.