My Math Forum Help ! roots of quartic equation.

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 December 15th, 2016, 01:30 PM #1 Newbie   Joined: Dec 2016 From: A place called UNIVERSE Posts: 8 Thanks: 0 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.
 December 15th, 2016, 02:23 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 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.
 December 15th, 2016, 05:11 PM #3 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 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).
 December 15th, 2016, 05:54 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Those are the correct values for m.
 December 17th, 2016, 06:47 AM #5 Newbie   Joined: Dec 2016 From: pakistan Posts: 2 Thanks: 0 to solve your problem.use synthetic division.

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