Try This Algebra Problem

$x^2=4$

Extract the square root of both sides.

$\sqrt{x^2}=\pm \sqrt4$

$x=\pm 2$

Therefore, $x=2$ and $x=-2$

Check:

when $x=2$ , we have

$2^2=4$

$4=4$ (true)

when $x=-2$

$(-2)^2=4$

$4=4$ (true)

Therefore, the solutions are

$x=2$ and $x=-2$ .

The solutions are $x = 2, -2$ .

We can factorize the equation as $0 = x^2 - 4 = (x-2)(x+2)$ , hence the solutions are $x = 2, -2$ .

The solutions are x=2,-2 We can substitute X by 2 and -2 in the given equation. Hence the solutions are x=2,-2.

$x ^ { 2 } = 4 \Leftrightarrow x = \pm 2$ by taking square root to each sides.

$x^2=4 \implies x=\pm 2$

$\large x^2 = 4$ , then

$\large x = 2, -2$

root of square of gives ans in +ve and -ve both