Question

how do you solve x^2+6x+9=1 by using the square root property?

Answer 1

\[x^2+6x+9 =(x+3)^2 \text{ \right?}\]

Answer 2

\[(x+3)^2=1 \]

Answer 3

now take square root of both sides
(x+3)=1 (x+3)=-1
x+3=1 x+3=-1

Answer 4

x=1-3 x=-1-3

Answer 5

x=-2 x=-4

Answer 6

Thankyou!

Answer 7

so for example in this problem, \[x ^{2}-8x+16=8\] it would be

Answer 8

\[(x-4)^{2}=8\]

Answer 9

right?

Answer 10

yes so far good! :)

Answer 11

and how do you get the last part again?

Answer 12

you want to isolate x
so first you need to isolate x-4
to do that you need to undo the square by square rooting both sides
but that gives us two equations
\[x-4=\pm \sqrt{8}\]

Answer 13

\[x-4=\sqrt{8} \text{ or } x-4=-\sqrt{8}\]

Answer 14

so you dont do the opposite ?

Answer 15

no tin these problems?

Answer 16

ok i'm going to do this problem and add one more step okay?

Answer 17

before i do that answer me this
do you know that
\[|x|=\sqrt{x^2}\]

Answer 18

yes.

Answer 19

ohhh okay so the asnwer is qoinq to be as a square root?

Answer 20

\[|x|=\pm x \text{ depending on if x is negative or positive }\]

Answer 21

ohhhh !

Answer 22

\[|-3|=-(-3)\]
\[|3|=+3\]

Answer 23

ok anyways lets do this problem one sec...

Answer 24

\[(x-4)^2=8\]
so this is the part you got to
you understand this step am i correct?

Answer 25

yes .

Answer 26

\[\sqrt{(x-4)^2}=\sqrt{8}\]
\[\text{ but } \sqrt{(x-4)^2}=|x-4|\]
\[\text{ so we actually have } |x-4|=\sqrt{8}\]

Answer 27

\[\text{ this means we have } x-4 = \pm \sqrt{8}\]

Answer 28

what about this?
is this good so far?

Answer 29

take your time and reading it

Answer 30

ohh so actually its square root of 8,

Answer 31

so in \[x ^{2}-2x+1=2\]

Answer 32

its \[(x-1)^{2}=2\]

Answer 33

then it would be \[x-1=\pm \sqrt{2}\]

Answer 34

yes yes

Answer 35

i got it ? :D

Answer 36

yep

Answer 37

\[x=1 \pm \sqrt{2} \text{ after we add 1 on both sides}\]

Answer 38

both sides ?

Answer 39

of the equal sign

Answer 40

we have x-1 on one side
and plus or minus square root of 2 on the other side

Answer 41

so it would be \[x=\pm \sqrt{2}+1\] ?

Answer 42

you can write it like that if you want

Answer 43

so on the previous one we didn't finish solving for x
we had \[x-4=\pm \sqrt{8}\]

Answer 44

then that one would be \[x=4\pm \sqrt{8}\]

Answer 45

yes and some teachers may want you to ''simplify" that

Answer 46

\[\sqrt{8}=\sqrt{4} \cdot \sqrt{2}=2 \sqrt{2}\]

Answer 47

so \[x=4 \pm 2 \sqrt{2}\]

Answer 48

good job itszel
i'm happy to help people who want to learn

Answer 49

so the answer would be \[x=4\pm2\sqrt{2}\] ?

Answer 50

yes for \[x^2-8x+16=8\]

Answer 51

thats the final answer?