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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 $\displaystyle ad$, $\displaystyle a$, $\displaystyle a+d$, $\displaystyle a+2d$. 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 relabel 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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