Question

Have an algebra question: solve by completing the square: x^2-4x+11 = 0
Can anyone help me with the answer?

Answer 1

yes lets to it the easy way

Answer 2

i'll check it at the end for you satellite73

Answer 3

\[x^2-4x=-11\]
\[(x-2)^2=-11+2^2=-11+4=-7\]

Answer 4

now if you are working with real numbers you can stop right there, because the left hand side is a prefect square and the right hand side is negativef

Answer 5

if you are working with comlex numbers you write
\[(x-2)^2=-7\]
\[x-2=\pm\sqrt{-7}\]
\[x-2=\pm\sqrt{7}i\]
\[x=2\pm\sqrt{7}i\]

Answer 6

so basically pw8589, what he's doing is isolating the x terms on one side and the constants on the other side. then divide you single cx (c standing for a constant) by 2 and square that number

Answer 7

I followed as x^2 -4x = -11
Then went to (x-2)^2 = -9
SR (x-2)^2 = +/- sr of -9
am I on the right track?

Answer 8

(x-2)^2 = -9 is not correct... you added 4 on the one side so you must add 4 on the other side.. you have -9, it should be -7

Answer 9

When taught completing the square - we divided the 4x by 2... which is where I was getting the 2 from instead of the 4

Answer 10

look at it this way

\[(x^2-4x)+4=-11 + 4\]

Answer 11

Thank you