$\displaystyle x^2=4$ $\displaystyle x=+/-2$ $\displaystyle x^2=y$ $\displaystyle x=sqrt(y)$ or $\displaystyle +/-sqrt(y)$ what is the ans and why?
Last edited by mr fantastic; Sep 4th 2010 at 02:35 PM. Reason: Re-titled.
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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 $\displaystyle x^2=y\iff x=\pm \sqrt{y}$
x^2=4y, x=?
Originally Posted by stupidguy x^2=4y, x=? Same thing. $\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$
$\displaystyle x=\pm \sqrt{4y}=\pm2\sqrt{y}$ if $\displaystyle x = \sqrt{4y^2} $ then $\displaystyle x = +2y $
Last edited by grgrsanjay; Sep 5th 2010 at 05:55 AM.
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?
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 $ .
Last edited by grgrsanjay; Sep 5th 2010 at 05:53 AM.
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.
i mean y^2 sorry
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 $\displaystyle x = \sqrt{4y^2} \implies x = 2|y|$
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?
Originally Posted by stupidguy Why? Think about when y is negative.
Originally Posted by stupidguy Why? because $\displaystyle \sqrt{x^2}= |x| $ is by the definition
if y is negative so be it.
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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