1. Square roots.

$\displaystyle x^2=4$
$\displaystyle x=+/-2$
$\displaystyle x^2=y$
$\displaystyle x=sqrt(y)$ or $\displaystyle +/-sqrt(y)$

what is the ans and why?

2. Originally Posted by stupidguy

$\displaystyle x^2=4$
$\displaystyle x=+/-2$
$\displaystyle x^2=y$
$\displaystyle x=sqrt y$ or $\displaystyle +/-sqrt y$

what is the ans and why?
(In $\displaystyle \mathbb{R}$)

$\displaystyle \sqrt{y}$ is defined nonnegative. In other words, $\displaystyle \sqrt{4} = 2$, not $\displaystyle \pm 2$.

Thus $\displaystyle x^2=y\iff x=\pm \sqrt{y}$

3. x^2=4y, x=?

4. Originally Posted by stupidguy
x^2=4y, x=?
Same thing.

$\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$

5. $\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$
if
$\displaystyle x = \sqrt{4y^2}$ then
$\displaystyle x = +2y$

6. Originally Posted by grgrsanjay
$\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$
if
$\displaystyle x = \sqrt{4y}$ then
$\displaystyle x = +2y$
you are confusing me!! wat u mean?

7. Originally Posted by grgrsanjay
$\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$ this is solution for your actual question
if
$\displaystyle x = \sqrt{4y^2}$ then if this is your question
$\displaystyle x = +2y$
.

8. Originally Posted by grgrsanjay
$\displaystyle x = \sqrt{4y}$ then
$\displaystyle x = +2y$
I also don't know what you mean. The statement

"$\displaystyle x = \sqrt{4y} \implies x = 2y$"

is only true if y = 1.

9. i mean y^2 sorry

10. Originally Posted by grgrsanjay
i mean y^2 sorry
But the statement

"$\displaystyle x = \sqrt{4y^2} \implies x = 2y$"

is still not generally true. It should be

$\displaystyle x = \sqrt{4y^2} \implies x = 2|y|$

11. Originally Posted by undefined
But the statement

"$\displaystyle x = \sqrt{4y^2} \implies x = 2y$"

is still not generally true. It should be

$\displaystyle x = \sqrt{4y^2} \implies x = 2|y|$
Why?

12. Originally Posted by stupidguy
Why?
Think about when y is negative.

13. Originally Posted by stupidguy
Why?
because $\displaystyle \sqrt{x^2}= |x|$ is by the definition

14. if y is negative so be it.

15. Originally Posted by yeKciM
because $\displaystyle \sqrt{x^2}= |x|$ is by the definition
Are you sure? Can someone verify? the absolute sign seems superfluous.

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