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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

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1 Answer

up vote 6 down vote accepted

You have:
$f(x)=x^6-5x^4+16x^2-72x+9$
$f'(x)=6x^5-20x^3+32x-72$
$f''(x)=30x^4-60x^2+32$

If you notice that $$f''(x)=30(x^4-2x^2+1)+2=30(x^2-1)^2+2 \ge 2 >0,$$ you can see that $f'(x)$ is strictly increasing. Together with $\lim\limits_{x\to-\infty} f'(x)=-\infty$ and $\lim\limits_{x\to\infty} f'(x)=\infty$ this implies that there is exactly one root $x_0$ of $f'(x)$.

Thus $f(x)$ is decreasing on $(-\infty,x_0)$ and increasing on $(x_0,\infty)$.

Since $\lim\limits_{x\to\infty} f(x)=\infty$ and $f(1)=1-5+16-72+9=-51$, we see that $f(x)$ has both positive and negative values.

Thus $f(x)$ must have two real roots, one of them in the interval $(-\infty,x_0)$ and another one in the interval $(x_0,\infty)$.


You can check the behavior of $f(x)$ and the behavior of $f'(x)$ at WolframAlpha.

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