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May 14th, 2015, 08:37 AM   #1
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Number of Solutions

Find the number of different real numbers x that satisfy (x^2+4x−2)^2 =(5x^2−1)^2
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May 14th, 2015, 08:58 AM   #2
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Originally Posted by ishaanmj007 View Post
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.
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May 14th, 2015, 09:23 AM   #3
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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.
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May 14th, 2015, 09:33 AM   #4
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Country Boy missed two cases.
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May 16th, 2015, 04:38 PM   #5
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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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