Prove that $x^{6} + x^{4} + x^{2} + x + 3 = 0$ has no positive real roots.

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Solution

$x^{6} + x^{4} + x^{2} + x + 3 = 0 \Rightarrow f (x)$

Using
Descartes method we can find maximum

No. of +ve or -ve roots

No. of +ve roots $=$ No. of sign changes. in $f (x)$

since $f (x)$
has no sign change

No +ve
real roots

$Eg: \quad
f(x)=2 x^{7}-x^{5}+4 x^{3}-5=0$

$f (x)$
changed sign three times

+ve to -ve to +ve to +ve to -ve

$3$ +ve real roots for (1)

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___

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Prove that x6+x4+x2+x+3=0 has no positive real roots.

Prove that $x^{6} + x^{4} + x^{2} + x + 3 = 0$ has no positive real roots.