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How to do I solve this problem? #3x^2-12=0#

I can't remember how to do these and just need step by step so I can finish the problem. This is only a portion of my calculus problem but I forget how to do this part. The whole problem is f(x) = x^3-12x+17 find the relative extreme points of the function if they exist. I did the first part and found the derivatives of the problem but now I am stuck at the next step which is what I posted for my question. It's been years since I took algebra so I forget. Thanks!

Answer 1

Hazel Anglin

#x = -2, x = 2#

You need to factor #3x^2-12#, which ends up being

#3(x+2) (x-2)#

and then set each of these (but ignore the #3#) equal to zero and solve.

#x+2=0#

#x+2- 2 = 0 - 2#

#x=-2#

and

#x-2=0#

#x-2+2=0+2#

#x=2#

So #x = 2, -2#

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Answer 2

Eli Assouline

To solve the equation (3x^2 - 12 = 0), you can follow these steps:

Add 12 to both sides to isolate the term with (x^2): (3x^2 = 12)
Divide both sides by 3 to solve for (x^2): (x^2 = \frac{12}{3})
Simplify: (x^2 = 4)
Take the square root of both sides: (x = \pm \sqrt{4})
Simplify: (x = \pm 2)

So, the solutions to the equation are (x = 2) and (x = -2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.