The number of real roots of (3-x)^4 + (5-x)^4 = 16 is?
Expand the first side and let what you get =16
the answer should be 2...
\[ (3-x)^4=x^4-12 x^3+54 x^2-108 x+81\\ (5-x)^4 =x^4-20 x^3+150 x^2-500 x+625\\ \] Now add both and set equal to 16
Actually x=3 is a solution.
and x=5 is also a solution.
There are two real solutions 3 and 5 and two complex solutions\[ 4 \pm i \sqrt 7 \]
\[ (3-x)^4+(5-x)^4-16=\\2 x^4-32 x^3+204 x^2-608 x+690=\\2 (x-5) (x-3) \left(x^2-8 x+23\right) \] Since 5 and 3 were easily found, you can divide by (x-5)(x-3) and get the factorization above.
The answer to the problem is 2.
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