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..

polynomials functional-equations
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edited Oct 21 "13 in ~ 3:47

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asked Oct 21 "13 in ~ 3:16

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

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^+\$

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
reply Oct 21 "13 at 3:37

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

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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