Solving 12x^2 - 9 = 0: What to Do Next?

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Homework Help Overview

The problem involves solving the quadratic equation 12x^2 - 9 = 0. Participants are exploring different methods to approach the solution without arriving at a final answer.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss factoring the equation, moving terms around, and the implications of squaring and square-rooting during the solving process. There are suggestions to try different methods if one does not yield results.

Discussion Status

The discussion is ongoing with various approaches being considered. Some participants provide guidance on the steps to take, while others express confusion about the operations involved in solving the equation. There is no explicit consensus on the best method yet.

Contextual Notes

Some participants question the application of operations to both sides of the equation and the implications of the square in the context of the problem. There are references to results from textbooks that differ from participants' calculations.

Jimmy84
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Homework Statement



12x^2 - 9 = 0

Homework Equations





The Attempt at a Solution



Im having doubts about how to start


I could factorize for 3(4x^2 - 3) = 0

or to pass -9 to the other side 12x^2 = 9 and keep solving.


What should I do?

Thanks.
 
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Just keep solving.
 
Jimmy84 said:
What should I do?
Pick one and try it.

If you can't solve the problem, pick the other one.

If you can't solve the problem, go back to the first one (or try finding a third option).
 
Mentallic said:
Just keep solving.

well the result on my book is - and + 1/2 square root of 3

If i keep solving I get 12x = +and- square root of 9 then x = +and- 3/ 12 then x = +and- 1/4 which is different.
 
do this step by step

first take the 9 over which gives:

12x^2 = 9

then what would u do?
 
sara_87 said:
do this step by step

first take the 9 over which gives:

12x^2 = 9

then what would u do?

I would move the square to the right, and I would have 12x = square root of nine isn't it?
 
When you make an operation in solving equations, you need to perform the operation on both sides, but not only that, you need to perform squaring, square-rooting, etc. for every factor.

e.g. If we have [tex]x^2=a^2b[/tex] then to solve for x you're obviously going to take the square root of both sides. This gives us [tex]x=\sqrt{a^2b}[/tex] (lets ignore the [itex]\pm[/itex] for the moment) and if you use the property of square roots, you'll have [tex]x=\sqrt{a^2}\sqrt{b}[/tex] and simplifying, [tex]x=a\sqrt{b}[/tex]

Now look at your problem: You didn't follow the above for when you took the square root of [tex]12x^2[/tex].
 
the ^2 only applies to the x

so remember it's 12 * x^2 = 9
 
take the 12 to the other side
 
  • #10
sara_87 said:
the ^2 only applies to the x

so remember it's 12 * x^2 = 9

you are right x^2= 9/12 then x^2 = 3/4 and then to include the square root.
x = square root of 3/4

= 1/2 square root of 3 thanks a lot guys.. I appreciate it .
 

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