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The equation $x^6 − 5x^4 + 16x^2 − 72x+ 9 = 0$ has

(A) exactly two distinct real roots

(B) exactly three distinct real roots

(C) exactly four distinct real roots

(D) six distinct real roots.

$f(0)>0$ and $f(1)<0$ and $f(3)>0$, so there should be an odd no of roots between 0 and 1 and 1 and 3 but exactly how many?

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marked as duplicate by Hagen von Eitzen, DonAntonio, Lord_Farin, Davide Giraudo, Git Gud May 4 '13 at 11:10

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2 Answers 2

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I haven’t tried it, but if you derive your equation twice, you can substitute $z = x^2$, solve for $z$ and get the inflection points of your equation from which you should be able to derive the number of zeros in the original equation.

Let’s see:

$$\frac{d^2}{dx^2}x^6−5x^4+16x^2−72x+9 = 30 x^4 - 60 x^2 + 32$$

Substituting $z = x^2$ and dividing by $2$, for the discriminant of $15z^2 - 30 z + 16$ you get: $900 - 4·15·16 = 900·(1 - \frac{32}{30}) < 0$, so you have no inflection points.

What does this tell us? Essentially that $x^6−5x^4+16x^2−72x+9 - y$ can only have two roots for if it had more than two, then it would have more than one extreme point and then it would have at least one inflection point by Rolle’s theorem.

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The second derivative of $f(x)=x^6 − 5x^4 + 16x^2 − 72x+ 9 = 0$ is $$f''(x)=30x^4-60x^2+32,$$ which is quadratic in $x^2$ having negative discriminant $-240$ and so is never zero. Since it's positive at $x=0$ we have $f''(x)>0$ for all real $x$, i.e. $f(x)$ is concave up on $\mathbb{R}$. This means $f(x)$ can have at most two real zeros, and you have already shown at least two.

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I can’t believe you beat me to 26 seconds. –  k.stm May 4 '13 at 10:35
    
It's not a race, your answer is essentially the same. I suppose lots of questions like this could be generated by taking irreducible quadratics, substituting $x^2$ for $x$, then integrating twice (or just adding a linear term which dies off at second derivative). –  coffeemath May 4 '13 at 10:48
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