Question

solve by completing the square :

x^2-6x-11=0

Answer 1

im there so far

Answer 2

no

Answer 3

if you could write out the process for solving it i could probably understand better

Answer 4

Oh, that was bad advice... I'll do the steps.

Answer 5

since 11 is prime the only two things to multiply is 1 and 11. adding/subtracting those in anyway will not make 6

Answer 6

What I meant was we want a form \[(x+a)^2 = b\]

Answer 7

Then to solve we will just have\[x=\sqrt{b}-a\]

Answer 8

quadratic formula

Answer 9

i know andres but they wont let me use it

Answer 10

\[(x+a)^2=x^2+2a+a^2\]In this case a needs to be -3

Answer 11

k

Answer 12

a is -3 because our equation has a -6x term. Now we need to reconcile the -11 term with some value for b.

Answer 13

sorry valpey but that isnt making sense for me. im kind of a visual person and need to see it step by step

Answer 14

\[(x+a)^2-b=0; \ and \ \ x^2-6x-11=0\]
\[x^2+2ax+a^2-b=0; \ and \ \ x^2-6x-11=0\]
so solve a^2-b = -11 knowing a is -3

Answer 15

ive been told that completing the square is just taking (b/2)^2 and adding it to both sides

Answer 16

\[x^2-6x-11=0\]add 11 to both sides
\[x^2-6x=11\]take half of 6 which is 3, square it get 9 and add 9 to both sides write
\[(x-3)^2=11+9=20\] take the square root, don't forget the \(\pm\)
\[x-3=\pm\sqrt{20}\] add 3 to both sides
\[x=3\pm\sqrt{20}\] you can write
\[x=3\pm2\sqrt{5}\] if you like

Answer 17

That is also true (but the "b" in that example is -6).

Answer 18

ahhh yes satellite thank you

Answer 19

to complete the square you divide the term with a single x by 2 and then square it. so wouldnt it be \[x ^{2}-6x-11=0\] then -6/2 =-3. then squared is +9.
the equation would then be \[x ^{2}-6x+9-9-11\] you add and then subtract to keep the equation the same, its like adding zero. you can factor that to \[(x-3)^{2}-20=0\]

Answer 20

this is just what i would do, sorry if its not what you need.

Answer 21

thanks all