SOLUTION: Totally lost with quadratic equations, can you please show me how to solve (x-4)^2 = -81 I have been through the book and notes several times and cannot figure out which met

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Totally lost with quadratic equations, can you please show me how to solve (x-4)^2 = -81 I have been through the book and notes several times and cannot figure out which met      Log On


   



Question 24534: Totally lost with quadratic equations, can you please show me how to solve
(x-4)^2 = -81
I have been through the book and notes several times and cannot figure out which method to use.
Thank you in advance for your help.

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
to solve the majority of quadratics, you want "equation equals zero" as your starting point, because you are being asked "where does the quadratic cross the x-axis...ie the y=0 line).

Now, your question is slightly different in the fact that you have a squared term on the left, so we can by-pass the whole "re-writing, factorising/quadratic formula" approach and just take the square root of both sides, to give:

x-4+=+sqrt%28-81%29 or x-4+=+-+sqrt%28-81%29

Now, the square root of a negative number is impossible to find, unless you have done Complex Numbers in maths. If you have not done them...then your answer is "no solution" ie the quadratic does not cross the x-axis.

If you have done Complex numbers, you get the following:

x-4+=+9i or x-4+=+-9i
--> x = 4 +- 9i

jon.