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I am given the polynomial

$$f = 2x^4-12x^2+2$$

and I have to show that all of the roots of this polynomial are real. I have no idea how to approach this. I tried plugging in the rational roots given by the rational root theorem, but none of them ($\pm1, \pm 2$) turned out to actually be roots, so I'm kind of lost.

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    It's a quadratic in $x^2$ – saulspatz Mar 22 '20 at 22:47
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    It’s a quadratic equation in $x^2$. You can find all of the roots directly, or just examine the discriminant. – amd Mar 22 '20 at 22:47
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    Notice that if you replace $x^2$ with $y$ you get a quadratic. From there you can find the roots explicitly. – TheMagicSnoot Mar 22 '20 at 22:48
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    We have $$2x^4-12x^2+2=2(x^4-6x^2+1)=2\big((x^2-1)^2-(2x)^2\big),.$$ You should be able to write all the roots in the form $\pm a\pm\sqrt{b}$, where $a$ and $b$ are positive integers. – Batominovski Mar 22 '20 at 22:53

6 Answers6

3

Let $t=x^2$, then the equation becomes: $2t^2-12t+2$. From here you have: $$x=\pm \sqrt{\frac{12\pm\sqrt{128}}{4}}$$ Because $12>\sqrt{128}$ all the roots are real.

Henno Brandsma
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Matteo
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2

HINT: Consider the polynomial $2y^2-12y+2$

Vasting
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We have $$\begin{align*} f(-3) &= +56, \\ f(-1) &= -8, \\ f(0) &= +2, \\ f(1) &= -8, \\ f(3) &= +56. \\ \end{align*}$$ Thus, $f$ changes sign four times, so all four roots of $f$ are real.

Eric M. Schmidt
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Factor it into a product of two quadrtatics, and check that the discriminants of those quadratics are positive.

So, $2(x^4-6x^2+1)=2(x^2-2x -1 )(x^2+2x -1 )$. The discriminants are $8$.

0

You can prove it without solving the biquadratic: this function has a local maximum and two local minima, and it tends to $+\infty$ at $\pm\infty$. Therefore, it suffices, by the intermediate value theorem, to show that the local maximum is nonnegative and the local minima are negative.

Indeed, $f'(x)=8x^3-24x)=8x(x^2-3)$ , so the critical values are $0, \sqrt 3,-\sqrt 3$. Also $f''(x)=24(x^2-1)$, which is negative on the interval $(-1,1)$, positive outside this interval. By the second derivative test, $0$ is the local maximum (and it equals $2$); the local minima are attained at $\pm\sqrt 3$ and $f(\pm\sqrt 3)=18-36+2=-16$.

Bernard
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Just find them all.

By quadratic equation

$x^2 = \frac {12\pm \sqrt{144- 16}}{4}$.

So the four solutions are

$\sqrt{\frac {12+ \sqrt{144- 16}}{4}}, -\sqrt{\frac {12+ \sqrt{144- 16}}{4}},\sqrt{\frac {12- \sqrt{144- 16}}{4}}, -\sqrt{\frac {12- \sqrt{144- 16}}{4}} $

Now as $144 > 164$ then $\sqrt{144-16}$ is real. And as $\sqrt{144-16} < \sqrt {144} = 12$ then $12 - \sqrt{144-16} > 0$ so $\sqrt{\frac {12- \sqrt{144- 16}}{4}}$ is real and all the roots are real.

fleablood
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