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Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

@mahmit2012 @timo86m @experimentX @stormfire1

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

negative roots?

OpenStudy (anonymous):

I am getting to it :P

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

just one

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

it has 2 negative roots

OpenStudy (anonymous):

they always come in pairs

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sqrt(-7/2)is complex

OpenStudy (anonymous):

The answer is 1

OpenStudy (anonymous):

they have one complex root

OpenStudy (anonymous):

this equation has 2 complex roots. 2 imaginary numbers

OpenStudy (anonymous):

one negative root and two complex

OpenStudy (anonymous):

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)]

OpenStudy (anonymous):

complex roots in real polynomials always come in pairs

OpenStudy (anonymous):

aaa ok :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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