What is the correct way to solve x^2=9 (see details)?

Here's how I teach it:

x² = 9
Take the square root of both sides:
√(x²) = √(9)
| x | = √(9)
| x | = 3
x = ±3

I remind the student beforehand that √(x²) is another way to write | x |.

x² = 9
x² = +/- 3

Factors:
= x + 3, x - 3

Answer: (x + 3)(x - 3) are the factors; root: 3, - 3

Original equation can also be stated as follows:
x² - 9 = 0

x^2 = 9
x = ±√9
x = ±√(3^2)
x = ±3

Check:
(-3)^2 = 9 ✔
3^2 = 9 ✔

I don't think that to write +/- x = 3 is "different" (better or worse) than to write x = +/- 3.

If you don't want to skip any step then you should write this:

|x| = |3|, which comes directly from one of the possible definition ways of absolute value of x.

From there you get directly to x = +/- 3.

Ana

It's not really a false impression. -3 and 3 are both square roots of 9! (+/- meaning plus OR minus, if that's where there's confusion)

3X3=9
-3X-3=9

Factoring does work as you showed, but the "direct" way is just the square root way with a step omitted (but it is still apparent that it's been employed).

It is important for people to know when to factor/use the quadratic formula, but it's also important for them to know when they can do what is simplest. Going on the easy route and taking the square root in one problem doesn't mean that they'll never factor and try to take square roots of negatives in other problems.

use the quadratic formula it's a more in depth version of what you're doing and you'll see that is how you get ± 3


x= -b+/- (square root of b to the second - 4ac) all divided by 2a

a=1
c=9


X = ± 3

x ² = 9
x = ± 3------------------end of story !

When you have x^2 = 9, the only thing stopping you from having a solution is that exponent. So to get rid of it and thus isolate x, it makes sense to take the square root of both sides of the equation to get x = ± √9, which means x = 3 or -3.

There's nothing wrong with doing it this way or doing it the following way:
x^2 = 9
x^2 - 9 = 0
(x+3)(x-3) = 0
x = -3, 3

But again, in the first case you're only one step away. So why not just get rid of the exponent and get the solution.

I have to laugh at the person who said to use the quadratic formula. Would somebody REALLY take x^2 = 9, and change it to x^2 - 9 = 0, and THEN plug a=1, b=0, c= -9 into the quadratic formula? It would work, but geez, what a waste of time.

x² = 9
x² -9 = 0
(x + 3)(x -3) = 0
x1 = +3
x2 = -3