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 $$x=\pm\sqrt{5-\sqrt{25-p}},\text{ when } 5\ge\sqrt{25-p}$$
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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, $p=9$.
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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. $$x^4-10x+9=(x^2-9)(x^2-1) = (x-3)(x+3)(x-1)(x+1)$$
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 $$ x^4-10x^2+p\ ? $$