1sir, is it x3 or x2 ?
if x2, then,
we have the given eqn. of the form
f(x)=f-1(x)
which is the same as solving f(x)=x, f(x) being x2+2ax+1/16,
or simply solving x2+(2a-1)x+1/16=0
to get the 2 roots..
cheers!!
Find the real roots of the equation:
x^3+2ax+\dfrac{1}{16}=-a+\sqrt{a^2+x-\dfrac{1}{16}}
where
0<a<\dfrac{1}{4}
Edit: The power of x is 2 and not 3 (thanx gordo for pointing it out).
And I placed it mistakenly in this section.
P.S: And yes gordo's solution (appearing in the next post) is correct.
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