Question 251347
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Looking at this it is possible that you meant that Question 1. is 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ -\ 24\ =\ 0]


In which case, divide by 6


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4\ =\ 0]


Now proceed in one of two ways:


<b>First method:</b> this is the difference of two squares, so factor it according to the pattern for the difference of two squares, to wit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)]


In your case, recalling that *[tex \Large 4\ =\ 2^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2^2\ =\ (x\ +\ 2)(x\ -\ 2)]


Then use the Zero Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2\ =\ 0\ \ \Rightarrow\ x\ =\ -2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 2\ =\ 0\ \ \Rightarrow\ x\ =\ 2]


<b>Second method:</b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4\ =\ 0]


Add 4 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 4]


Take the square root of both sides, remembering to consider both the positive and negative roots:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x^2}\ =\ \pm\sqrt{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm2]


Same answer either way.  Next time be more careful about how you render your question so that we don't have to guess at what you mean.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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