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 34173This 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?
Thanks! This question is from textbook Algebra Beginning and Intermediate
You can put this solution on YOUR website! 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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