Question

HELP ME PLEASE ALGEBRA 2!

Answer 1

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