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December 15th, 2016, 01:30 PM   #1
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Help ! roots of quartic equation.

Need assistance in following problem:

Find all values of m for which the quartic equation x^4 - (3m+ 2)x^2 +m^2 = 0
has 4 real roots in arithmetic progression.
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December 15th, 2016, 02:23 PM   #2
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Hint: (x - a)(x - a - d)(x - a - 2d)(x - a - 3d)
= x^4 - (4a + 6d)x³ + (6a² + 18ad + 11d²)x² - (4a + 6d)(a² + 3ad + d²)x + a(a + d)(a + 2d)(a + 3d).

Substituting a = -3d/2 reduces that to x^4 - (5/2)d²x² + (9/16)d^4.
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December 15th, 2016, 05:11 PM   #3
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Let the roots be $\displaystyle a-d$, $\displaystyle a$, $\displaystyle a+d$, $\displaystyle a+2d$. Now google elementary symmetric functions. Among other things, the sum of the roots is zero. So we get $\displaystyle 4a+2d=0$ and $\displaystyle d=-2a$. So we can re-label the roots as $\displaystyle -3a,-a,a,3a$. I think if you continue, you get $\displaystyle m=6$ or $\displaystyle m=\dfrac{-6}{19}$ (but I haven't checked the answers).
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December 15th, 2016, 05:54 PM   #4
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Those are the correct values for m.
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December 17th, 2016, 06:47 AM   #5
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to solve your problem.use synthetic division.
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