Question

How do I find the solution of x^2 - 81 using square roots? I need help >.<

Answer 1

Start by getting \(x^2\) by itself

Answer 2

Isn't it by itself already?

Answer 3

Nope, there's a -81 next to it o:

Answer 4

\[-81x^2\]?

Answer 5

@Luigi0210

Answer 6

No.. like this.
\[\LARGE x^2-81=0\]
Get x^2 by itself
\[\LARGE x^2=81\]
Now take the sqaure root of both sides, what do you get?

Answer 7

The square root of 81 is 9

Answer 8

I'm not sure what the square root of \[x^2\]

Answer 9

\[\sqrt{x ^{2}} \]
is just x because they cancels out

Answer 10

What do I do next?

Answer 11

you already got the answer x=9

Answer 12

Well... My answer choices are -9(squared) and 9(squared)
-81 and 81
-9 and 9
no real number solutions

Answer 13

Is it option C.?

Answer 14

Not squared sorry, I meant square rooted in the parenthesis.

Answer 15

yes if you meant -9 and 9

Answer 16

Okay thanks :)

Answer 17

as \[\sqrt{81} = \pm9\]

Answer 18

Could you help me with one more?

Answer 19

sure

Answer 20

\[3x^2-147\]

Answer 21

Isolate the x?

Answer 22

yes

Answer 23

\[x^2=147(3)\]?

Answer 24

uhh nope

Answer 25

you isolate by subtracting 147 first\[3x ^{2} =147\]

Answer 26

Subtracting 147-3?

Answer 27

to bring the 3 to the other side, you do the opposite
what is the opposite of multiplying by 3?

Answer 28

Dividing, okay so 147/3=49

Answer 29

Then square root both so the square root of 49 is 7

Answer 30

Ohhhh it's easy :D I'm just over complicating things.

Answer 31

yes! but careful \[\sqrt{49} = \pm 7 \]

Answer 32

What do you mean be careful?

Answer 33

usually a question asking for sqrt of 49 would be 7, but then it could also be -7 at the same time as -7x-7 also equals 49

Answer 34

Yeah but I'm okay because the answer choice is -7 and 7.

Answer 35

yes correct

Answer 36

Yayy, thank you for your help!

Answer 37

no problem