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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. Tags equation, quartic, roots Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post goldgold Algebra 5 December 26th, 2014 05:15 AM Jopus Algebra 0 May 23rd, 2014 10:38 AM hoyy1kolko Algebra 6 October 28th, 2011 12:54 AM ally Algebra 2 July 24th, 2008 11:51 AM rnck Abstract Algebra 0 December 31st, 1969 04:00 PM

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