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In my previous question I asked what is $\sqrt{0}$ and learned that it is 0. I learned also that $\sqrt{2}^2=2$ in my class. Then I see $\sqrt{0}^2=0^2=0$. So in general to find $\sqrt{x}$ is it to find $y$ such that $y^2=x$?

Also i see it works for $\sqrt{4}=2$ because $\sqrt{4}^2=2^2=4$. So $\sqrt{4}$ is $y$ such that $y^2=4$.

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Jacob
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1 Answers1

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It is true that $y = \sqrt{x}$ will satisfy $y^2 = x$. For real numbers, as long as $x \geq 0$ then $x$ has a real square root. The only thing you have to add to your explanation is that you actually find two solutions to $y^2 = x$ (except for $x=0$).

For example, as you say $\sqrt{2}^2 = 2$ but also $(-\sqrt{2})^2 = 2$. This is why you will often see that when people solve $y^2 = x$ they will write $y = \pm \sqrt{x}$, meaning that $y$ could be either of those values and satisfy the equation.

To summarize, we choose to write $\sqrt{x}$ as the positive solution $y$ to $y^2=x$, for $x >0$.

Jason Knapp
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  • Thanks. So is there a link between the square in $ax^2+bx+c$ and the $\pm$ in $(-b\pm\sqrt{b^2-4ac})/2a$? Is it for the same reason? – Jacob Dec 12 '14 at 21:09
  • Because I just realize that finding $\sqrt{x}$ is like $ay^2+by+c$ but with $a=1$, $b=0$ and $c=0$. – Jacob Dec 12 '14 at 21:10
  • Does this make sence? – Jacob Dec 12 '14 at 21:11
  • Yep. If you look at a derivation of the quadratic formula, for example on Wikipedia (http://en.wikipedia.org/wiki/Quadratic_formula), you will see a point where they solve $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$, and when you take square roots at this point you introduce the ambiguity. – Jason Knapp Dec 12 '14 at 21:12
  • No it is $c=-x$ in fact. – Jacob Dec 12 '14 at 21:12
  • Yes the formula gives $y=\pm\sqrt{x}$ so it works like your answer. – Jacob Dec 12 '14 at 21:13