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Problem :

Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly four distinct real roots, then which of the following options are correct :

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

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

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

(d) none of these

My approach :

Since $f(x^2)=0 $ has exactly four distinct real roots. Therefore the remaining two roots left [ as $ f(x^2)=0$ is a degree six polynomial].

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

So, option (c) is incorrect. I think the answer lies in option (a) or (b) . But I am not confirm which one is correct. Please suggest.. thanks..

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  • $\begingroup$ "Four distinct real roots" could mean that there are 4 real roots and 2 complex roots, or it could mean that there are 4 real roots, 0 complex roots, but some of the real roots have multiplicity. You will have to clarify the question to get a correct answer. I suggest checking your textbook for context. $\endgroup$ – DanielV Oct 21 '13 at 4:25
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No requirement for $f$ to have real coefficients was stated. So you could have e.g. $f(x) = (x-1)(x-4)(x+i)$ which has two real roots, and $f(x^2)$ has the four distinct real roots $\pm 1$ and $\pm 2$.

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I'm editing this because found a fatal error on the demonstration assuming $C\in\mathbb{R}$ witch may not be always true, leading to a wrong answer. Excuse me :/

First condition: $f(x)$ is 3rd degree. So, $f(x)$ can be expressed as $(x-A)(x-B)(x-C)$ with $A,B,C \in \mathbb{C}$.

*Note that if $A \in \mathbb{C}$, it's possible that $A\in\mathbb{R}$.

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

Second condition: $f(x^2)$ has exactly four distinct real roots. It means that $A \ge 0, A\ne B\ge0, C \in (\mathbb{C}-\mathbb{R^+})$. I'm Assuming $0 \in \mathbb{R^+}$

Answer: d)

Cannot be a) because if $C \in \mathbb{C-R}$, $x=C$ is complex, so $f(x)$ does not have 3 real roots.

Cannot be b) because if $C \in \mathbb{R}$, $x=C$ is real, witch is the third root.

Cannot be c) because $A$ and $B$ are real roots of the equation.

Is d) because it says it cannot be a) b) and c).

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  • $\begingroup$ Shouldn't it should be $x=\pm\sqrt{a,b}$? $\endgroup$ – chubakueno Oct 21 '13 at 3:56
  • $\begingroup$ True that, thanks :p $\endgroup$ – Alan Oct 21 '13 at 4:27
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Both $f(x)=(x-1)(x-4)^2$ (two real roots including one double root), and $f(x)=(x+1)(x-1)(x-4)$ (three real roots) fulfit the conditions of the problem, with $f(x^2)=0$ having roots in $x=\pm1$ and $x=\pm2$.

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  • $\begingroup$ The first example could be called "three real roots counted by multiplicity". $\endgroup$ – Robert Israel Oct 21 '13 at 3:43
  • $\begingroup$ And that would mean that $f(x^2)=0$ has six real roots counted by multiplicity. What the OP meant by distinct real roots (if it includes or excludes multiplicity) I cannot tell. $\endgroup$ – Carlos Eugenio Thompson Pinzón Oct 21 '13 at 3:52
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    $\begingroup$ Distinct means not counted by multiplicity. $\endgroup$ – Robert Israel Oct 21 '13 at 5:52
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I assume that the polynomial has real coefficients.

Now if $f(x^2)$ has four distinct real roots, then the square of these roots are the roots of $f(x)$. Hence the polynomial has at least two distinct real roots. Hence it has three distinct real roots (as complex roots occur in pairs).

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