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

@mahmit2012 @timo86m @experimentX @stormfire1

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

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

negative roots?

I am getting to it :P

ok

just one

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

first (x^3+3)=0 x^3=-3 x=cuperoot(-3) ------ second (2*x^2+7)=0 2x^2=-7 x^2=-7/2 x=sqrt(-7/2)

it has 2 negative roots

they always come in pairs

Raph dont they always come in pairs? so it is 2?

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

sqrt(-7/2)is complex

The answer is 1

they have one complex root

this equation has 2 complex roots. 2 imaginary numbers

one negative root and two complex

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)]

complex roots in real polynomials always come in pairs

aaa ok :)

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

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

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

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

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