Let $f(x)$ be a third degree polynomial such that $f(x^2)=0$ has precisely four unique real roots, climate which that the following choices are correct :

(a) $f(x) =0$ has actually all three real roots

(b) $f(x) =0$ has exactly two genuine roots

(c) $f(x) =0$ has only one real root

(d) nobody of this

My method :

Since $f(x^2)=0 $ has specifically four unique real roots. As such the staying two roots left < together $ f(x^2)=0$ is a level six polynomial>.

You are watching: If f(x) is a third degree polynomial function, how many distinct complex roots are possible?

How can we say that the continuing to be two roots will be genuine . This will not have one genuine root ( as non real roots come in conjugate pairs).

So, option (c) is incorrect. I think the price lies in choice (a) or (b) . Yet I am no confirm i beg your pardon one is correct. Please suggest.. Thanks..


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edited Oct 21 "13 in ~ 3:47
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SachinSachin
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No necessity for $f$ to have real coefficients was stated. For this reason you could havee.g. $f(x) = (x-1)(x-4)(x+i)$ which has two actual roots, and $f(x^2)$ has the four unique real root $pm 1$ and also $pm 2$.


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answer Oct 21 "13 in ~ 3:47
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Robert IsraelRobert Israel
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I"m modifying this due to the fact that found a fatal error top top the demonstration suspect $Cinptcouncil.netbbR$ witch may not be always true, bring about a not correct answer. Forgive me :/

First condition: $f(x)$ is 3rd degree. So, $f(x)$ deserve to be expressed together $(x-A)(x-B)(x-C)$ v $A,B,C in ptcouncil.netbbC$.

*Note that if $A in ptcouncil.netbbC$, it"s possible that $Ainptcouncil.netbbR$.

If $f(x)=(x-A)(x-B)(x-C)$ climate $f(x^2)=(x^2-A)(x^2-B)(x^2-C)$

Second condition: $f(x^2)$ has exactly four distinct real roots. It method that $A ge 0, A e Bge0, C in (ptcouncil.netbbC-ptcouncil.netbbR^+)$. I"m suspect $0 in ptcouncil.netbbR^+$

Answer: d)

Cannot be a) because if $C in ptcouncil.netbbC-R$, $x=C$ is complex, so $f(x)$ walk not have actually 3 real roots.

Cannot it is in b) since if $C in ptcouncil.netbbR$, $x=C$ is real, witch is the 3rd root.

Cannot it is in c) since $A$ and also $B$ are genuine roots of the equation.

Is d) due to the fact that it states it cannot be a) b) and also c).


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edited Oct 21 "13 at 21:02
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AlanAlan
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Both $f(x)=(x-1)(x-4)^2$ (two actual roots consisting of one twin root), and also $f(x)=(x+1)(x-1)(x-4)$ (three actual roots) fulfit the conditions of the problem, through $f(x^2)=0$ having actually roots in $x=pm1$ and $x=pm2$.


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answer Oct 21 "13 in ~ 3:22
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Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
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I assume the the polynomial has actually real coefficients.

Now if $f(x^2)$ has actually four distinct real roots, then the square of this roots space the root of $f(x)$. Hence the polynomial has at least two distinctive real roots. Hence it has three unique real root (as complicated roots take place in pairs).


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reply Oct 21 "13 at 4:47
Vishal GuptaVishal Gupta
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