Hello from MrBillDoesMath!
Answer:
Case 1: x is a real number. In this case f(x) has no real solution.
Case 2: x is a complex number. In this case the roots of f(x) are
x = 3 +\- 2 sqrt(3) i
(sqrt = square root. "i" = sqrt(-1))
Discussion:
Add 1 to both sides of the given equation:
2(x-3)^2 -1 + 1 = -25 + 1 or
2(x-3)^2 + 0 = -24 or
2(x-3)^2 = -24 (*)
Case 1: x is a real number. In that case as the square (x-3)^2 is always >0, so (*) has no real solution.
Case 2. x is a complex number ( a + bi, where a <> 0). In this case divide both sides of (*) by 2,
(x-3)^2 = -24/2 = -12 =>
x-3 = +\- sqrt(-12) = +\- sqrt (12 * -1) = +\- sqrt (4 * 3 * -1) = +\- 2 sqrt(3) * i
where i = sqrt (-1).
Add 3 to both sides gives
x = 3 +\- 2 sqrt(3) i
Thank you,
MrB
