SOLUTION: I need help with the following 2 questions... 1. Solve by extracing square roots: 3x^2-80=0 I figured this was it, can you check the work? 3x^2

SOLUTION: I need help with the following 2 questions... 1. Solve by extracing square roots: 3x^2-80=0 I figured this was it, can you check the work? 3x^2 - 80 = 0 3x^2 = 80 3x^2/3 = 8

Question 34172This question is from textbook Algebra Beginning and Intermediate
: I need help with the following 2 questions...
1. Solve by extracing square roots: 3x^2-80=0
I figured this was it, can you check the work?
3x^2 - 80 = 0
3x^2 = 80
3x^2/3 = 80/3
x^2 = 80/3
sq root of x^2 = sq root of 80/3
Answer - x=sq root of (= or -)80/3?? Does the (+ or -) have to be there?
2. Solve by extracting square roots: (2x-2)^2 = 9
sq root of (2x-2)^2 = sq root of 9
2x-2 = (+ or -)3
This is where I am confused... is the next step dividing it all by 2, or taking the -2 to the other side of the problem? Could you please help me get though this one?
This question is from textbook Algebra Beginning and Intermediate

Answer by Cintchr(481) (Show Source): You can put this solution on YOUR website!
This is a duplicate of Problem number 34173

Not too bad ... you just need to keep going ...
Problem 1:

now when you square root the fraction .. you will get a denominator with a square root in it ... That is a BIG NO-NO!!!

To rationalize the denominator we need to multiply the numerator and denominator by root3

And yes ... it is +-

Done.

Problem 2:

yes ... root both sides
here you will have two roots... one positive ... one negative
or
ADD 2
or
Divide by 2
or
When solving for a given variable .... we work PEMDAS backwards ... doing the adding and subtracting first ... and then the multiplying and dividing

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