MHB Can a Polynomial Equation Have Roots in Arithmetic Progression?

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The discussion centers on finding values for $a$ and $b$ in the polynomial equation $x^6 + ax^4 + bx^2 - 225 = 0$ such that it has six real roots in arithmetic progression. Participants share their solutions and approaches to the problem, with several members successfully identifying the correct values. The solutions involve manipulating the polynomial and applying properties of roots in arithmetic progression. Key insights include the relationship between the coefficients and the roots' properties. The thread highlights collaborative problem-solving in advanced polynomial equations.
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Find $a$ and $b$ such that the equation $x^6+ax^4+bx^2-225=0$ has six real roots in arithmetic progression.


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Congratulations to the following members for their correct solutions::)

1. MarkFL
2. kaliprasad
3. greg1313
4. Fallen Angel
5. lfdahl
6. chisigma

Solution from kaliprasad:
Because of coefficient of $x^5 = 0$, hence the sum of the six roots is zero.

Let the roots be $-5a,-3a,-a,a,3a,5a$ with $a \gt 0$.

$\text{product of roots} = -225a^6 = -225$ or $a= 1$

so roots are -5,-3,-1,1,3,5.

Given the polynomial is

$(x+5)(x+3)(x+1)(x-1)(x-3)(x-5) = (x^2-25)(x^2-9)(x^2-1)$

=$(x^6 - (25+9+1)x^4 + ( 25 * 9 + 25 * 1 +9) x^2 - 225$

= $x^6 - 35x^4 + 259x^2-225$

so a = - 35 and b = 259

Solution from Fallen Angel:
First of all, we can change $y=x^2$ and get the polynomial $P(y)=y^3+a^2+by-225$

Now if $\alpha ,\beta ,\gamma$ are the roots of $P$, the condition of the roots of the original polynomial being in arithmetic progression implies $\sqrt{\beta}=3\sqrt{\alpha}$ and $\sqrt{\gamma}=5\sqrt{\alpha}$

Now we can factor $P(y)=(y-\alpha)(y-9\alpha)(y-25\alpha)$ and we obtain $\alpha=1$, $a=-35$, $b=259$.

Back to the original polynomial we got that the roots of $x^6-35x^4+259x^2-225$ are $-5,-3,-1,1,3,5$.

Solution from lfdahl:
Given $P(x) = x^6+ax^4+bx^2-225 = 0$.

$P(x) = P(-x)$ so $P$ is an even function. Therefore, the roots are symmetrically distributed around $x = 0$.

If I let $r$ denote the smallest positive root (and $-r$ the largest negative) I need to factorize $P(x)$ such that the numerical distance between neighbouring roots is $2r$: $\pm r, \pm 3r, \pm 5r$, and $P(x)$ has the form:

$P(x) = (x+r)(x-r)(x+3r)(x-3r)(x+5r)(x-5r)= (x^2-r^2)(x^2-9r^2)(x^2-25r^2) $.

The root product $(-r^2)(-9r^2)(-25r^2)$ should be $-225$: $-9\cdot25r^6 = -225$ so $r = \pm 1$.

Thus the factorization is simply: $P(x) = (x^2-1)(x^2-9)(x^2-25) = x^6-35x^4+259x^2-225$.

The coefficients $a$ and $b$ must therefore be: $a=-35$ and $b = 259$.
 
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