Doubt about solving a simple quadratic equation

[issacnewton]

Homework Statement

Solve $x^2 = 4$

Relevant Equations

rules of factoring and absolute values

I was thinking of this simple equation here, $x^2 = 4$. Many students present the solution as follows.
$$ x^2 = 4 $$
$$ \therefore x = \sqrt{4} = \pm 2 $$
Now, even though the final answer is correct, there is a mistake in arriving at the solution. Square root symbol means that we have to take positive square root only. Following is a correct method in my opinion.
$$ x^2 = 4 $$
$$ \therefore |x|^2 = |2|^2 $$
$$ \sqrt{|x|^2} = \sqrt{|2|^2}$$
Now, since, $|y| = \sqrt{y^2}$ for any $y$, we have
$$ ||x|| = ||2|| $$
$$ |x| = |2| = 2 $$
Now, either $x \geq 0$ or $x < 0$, so, we get two solutions. $x = 2$ and $- x = 2$. So, finally, we have $x= 2$ or $x = -2$
I think this would be rigorous way of solving this. I myself was confused about this for a while. How do you see students solving such an equation ?

 

[etotheipi]

If you have an equation like $x^{2} = 4$, the way I'd do it "formally" is to take the principal square root of both sides:

$\sqrt{x^{2}} = \sqrt{4}$

and then use the definition of the principal square root that you alluded to, $\sqrt{k^{2}} = |k|$,

$|x| = 2$

And this leads to $x=\pm2$.

 

[PeroK]

Homework Statement:: Solve $x^2 = 4$
Homework Equations:: rules of factoring and absolute values

I was thinking of this simple equation here, $x^2 = 4$. Many students present the solution as follows.
$$ x^2 = 4 $$
$$ \therefore x = \sqrt{4} = \pm 2 $$

$$ x^2 = 4 $$
$$ \therefore x = \pm \sqrt{4} = \pm 2 $$

 

[mjc123]

Or x² = 4
x² - 4 = 0
(x+2)(x-2) = 0
x+2 = 0 or x-2 = 0
x = 2 or -2.

 

[issacnewton]

Yes, factoring is another way to solve this. But, I see lot of students confused about the use of square root. So, while teaching, it would be helpful to tell about the formal approach.

 

[SammyS]

Yes, factoring is another way to solve this. But, I see lot of students confused about the use of square root. So, while teaching, it would be helpful to tell about the formal approach.

Technically speaking, $\sqrt{x^2\ } = |x|$.

 

[issacnewton]

Thanks