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 May 14th, 2015, 08:37 AM #1 Newbie   Joined: May 2015 From: India Posts: 14 Thanks: 0 Math Focus: Number Theory Number of Solutions Find the number of different real numbers x that satisfy (x^2+4x−2)^2 =(5x^2−1)^2
May 14th, 2015, 08:58 AM   #2
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Quote:
 Originally Posted by ishaanmj007 Find the number of different real numbers x that satisfy (x^2+4x−2)^2 =(5x^2−1)^2
This reduces immediately to x^2+ 4x- 2= 5x^2- 1 and x^2+ 4x- 2= -(5x^2- 1)= 1- 5x^2.

The first, x^2+ 4x- 2= 5x^2- 1 is the same as 4x^2- 4x+ 1= 0 and x^2+ 4x= 1- 5x^2 is the same 6x^2+ 4x- 1= 0.

 May 14th, 2015, 09:23 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This is asking for the number of distinct real roots of 24x^4 - 8x^3 - 22x^2 + 16x - 3. You can factor this polynomial as $$(ax+b)^2(cx^2+dx+e)$$ and check whether the discriminant of the latter is positive, negative, or 0. Thanks from topsquark and ishaanmj007
 May 14th, 2015, 09:33 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Country Boy missed two cases.
May 16th, 2015, 04:38 PM   #5
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Quote:
 Originally Posted by greg1313 Country Boy missed two cases.
??

Either both quadratics will be the same sign, or they will be the opposite sign.

Country Boy's first resultant quadratic equation gives the repeated x = 1/2.

His second resultant quadratic equation gives two distinct irrational values
for x.

Country Boy's two resultant quadratic equations will give the same number
of solutions as the fourth degree equation posted above by CRGreathouse.

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