Question

What is the solution of the equation
x^2-2x=22

a. x=1+sqrt21
b. x=1+sqrt23
c. x=-1+sqrt21
d. x=-1+sqrt23

Answer 1

easiest here is to complete the square
do you know how to do it?

Answer 2

if the answer is "no" that is fine, i can show you

Answer 3

Ok thanks

Answer 4

what is half of \(2\)?

Answer 5

1

Answer 6

right, and \(1^2=1\) so start with
\[(x-1)^2=22+1\]

Answer 7

ok then x-1=22

Answer 8

then since \(22+1=23\) you have
\[(x-1)^2=23\]
take the square root of both sides, then add 1 to both sides and you are done

Answer 9

oh ok

Answer 10

so the sqrt is 23

Answer 11

x=1+sqrt23?

Answer 12

yes \[x-1=\sqrt{23}\]

Answer 13

negative?

Answer 14

so \(x=1+\sqrt{23}\) right

Answer 15

ok thanks :)

Answer 16

can you help me with another 1?

Answer 17

sure

Answer 18

x^2+6x+41=0

a. x=-3+1 sqrt41
b. x=-3+4i sqrt2
c. x=-3+5sqrt2
d. x=3+4i sqrt 2

Answer 19

lets to this one the same way
start with \[x^2+6x=-41\] then complete the square
what is half of \(6\)?

Answer 20

3

Answer 21

and \(3^2\)?

Answer 22

9

Answer 23

ok
then we go right from
\[x^2+6x=-41\] so
\[(x+3)^2=-41+9\] what is \[-41+9\]?

Answer 24

-32

Answer 25

right so
\[(x+3)^2=-32\] making
\[x+3=\pm\sqrt{-32}\]

Answer 26

ok..

Answer 27

subtract \(3\) from both sides

Answer 28

do you know how to rewrite \(\sqrt{-32}\) using \(i\)?

Answer 29

no..sorry :I

Answer 30

ok lets do that

Answer 31

ok

Answer 32

\[-32=16\times 2\times -1\]si
\[\sqrt{-32}=\sqrt{16\times 2\times -1}=\sqrt{16}\sqrt2\sqrt{-1}=4\sqrt2i\]

Answer 33

ok so we know its 4sqrt2i?

Answer 34

or 4isqrt2

Answer 35

right
final answer
\[x+3=4\sqrt2 i\\
x=-3+4\sqrt2 i\]

Answer 36

so itts b

Answer 37

yeah they put the \(i\) in the middle because they don't know what they are doing

Answer 38

but B is right

Answer 39

lol ok thanks

Answer 40

yw