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What's the value of $p$ if the roots of the biquadratic equation $$x^4-10x^2+p=0$$ are in AP?

The given equation is quadratic in $x^2$, so it's discriminant is $D=25-p\ge0\iff p\le25$ and the roots are $\left(x^2\right)_{1,2}=5\pm\sqrt{25-p}$. For $x$ we have $$x=\pm\sqrt{5+\sqrt{25-p}}$$ and $$x=\pm\sqrt{5-\sqrt{25-p}},\text{ when } 5\ge\sqrt{25-p}$$

yinivem462
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3 Answers3

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Hint

If the roots are $\ a-3d, a-d, a+d, a+3d\ $ (any arithmetic progression can be written in this form), what values of $\ a\ $ and $\ d\ $ will reduce the polynomial $$ (x-a+3d)(x-a+d)(x-a-d)(x-a-3d) $$ to the form $$ x^4-10x^2+p\ ? $$

lonza leggiera
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  • Thank you for the response! Why are we assuming that the equation has 4 real roots, though? – yinivem462 Apr 14 '23 at 15:27
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    You don't need to assume that all the roots are real. If you recover all sets roots that are in arithmetic progression, including the complex ones, you'll also have all sets of real roots that are in arithmetic progression. (I'm not assuming that $\ a\ $ and $\ d\ $ are real. – lonza leggiera Apr 14 '23 at 15:33
  • May I ask you why did you decide to write the AP exactly as $a-3d, a-d, a+d, a+3d$? – yinivem462 Apr 14 '23 at 18:47
  • I had initially taken the roots to be $\ a,$$,a+d,$$,a+2d,$$,a+3d\ $. But when I wrote down the product $\ (x-a)$$(x-a-d)$$(x-a-2d)$$(x-a-3d)\ $, I noticed that if I doubled the difference and shifted $\ a\ $ down by $\ 3d\ $ I'd be able to simplify the calculation by pairing up the factors $\ x-a-d\ $ and $\ x-a+d\ $ and $\ x-a-3d\ $ and $\ x-a+3d\ $ – lonza leggiera Apr 14 '23 at 22:03
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Based on symmetry, we can write the four roots as $a\pm b$, and $a\pm 3b$, where the common difference is equal to $d=2b$. Then we have \begin{align} (x-(a+3b))(x-(a+b))(x-(a-b))(x-(a-3b))&=0\\ (x^2-2ax+a^2-b^2)(x^2-2ax+a^2-9b^2)&=0\\ x^4-4ax^3+(6a^2-10b^2)x^2+(20ab^2-4a^3)x+(a^4-10a^2b^2+9b^4)&=0 \end{align} It then follows that $$\left\{\begin{align} -4a&=0\\ 6a^2-10b^2&=-10\\ a^4-10a^2b^2+9b^4&=p \end{align}\right.$$

We can see trivially that $a=0$, $b=\pm1$, and subsequently, $p=9$.

Andrew Chin
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Now that you have the four roots (and condition on $p$ for these to be real), you could put them in order $$-\sqrt{5+\sqrt{25-p}}$$ $$-\sqrt{5-\sqrt{25-p}}$$ $$\sqrt{5-\sqrt{25-p}}$$ $$\sqrt{5+\sqrt{25-p}}$$ Since they have to be in an arithmetic progression, the difference between two consecutive roots must always be the same. We are able to create an equation for $p$

The difference between the third and second root is $$2\sqrt{5-\sqrt{25-p}}$$ The difference between the fourth and third root is $$\sqrt{5+\sqrt{25-p}}-\sqrt{5-\sqrt{25-p}}$$ These two differences must be the same $$\sqrt{5+\sqrt{25-p}}-\sqrt{5-\sqrt{25-p}} = 2\sqrt{5-\sqrt{25-p}}$$ Then solving for $p$ $$\sqrt{5+\sqrt{25-p}} = 3\sqrt{5-\sqrt{25-p}}$$ $$5+\sqrt{25-p} = 9\left(5-\sqrt{25-p}\right)$$ $$\sqrt{25-p} = 4$$ $$p = 9$$ Finally, try this in your polynomial to make sure it works. $$x^4-10x+9=(x^2-9)(x^2-1) = (x-3)(x+3)(x-1)(x+1)$$

Alain Remillard
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