2
$\begingroup$

So if you take $x=y^2$ and get the sqrt of both sides you get $y=\sqrt{x}$ so they are the same right? But when you graph them, $y=\sqrt{x}$ only shows the positive $y$ values because you can't sqrt a negative number because no 2 same numbers multiplied together are negative and what not. And of course $x=y^2$ shows both positive and negative $y$ values. So why does this happen with this?

EDIT: My girlfriend pointed out that in $x=y^2$ the $x$ technically cannot be negative cuz if the $y^2$ will never be negative. But why do graphs show the negatives? And I forgot to show you the site where it graphs the negatives: http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/09-sideways-parabolas-01

$\endgroup$
4
$\begingroup$

Simply because $x=y^2$ doesn't imply that $y=\sqrt x$.

$\endgroup$
  • 1
    $\begingroup$ So does x=y^2 not mean the same as y=sqrt(x) but actually mean y=+/-sqrt(x)? $\endgroup$ – gopro_2027 Mar 22 '15 at 4:07
  • $\begingroup$ Yes, it's exactly that ! $\endgroup$ – Surb Mar 22 '15 at 4:07
  • $\begingroup$ @gopro_2027 Similar situations occur for $x=y^3$ and $x=y^n$ in general. You have $\sqrt[n]{x}=|y|$, but more specifically $\sqrt[n]{x}=\zeta_n^k y$ where $\zeta_n^k$ is some $n^{th}$ root of unity raised to some integer power. For $x=y^4$ you have $\sqrt[4]{x}\in\{y,iy,-y,-iy\}$ $\endgroup$ – JMoravitz Mar 22 '15 at 4:14
  • $\begingroup$ $y=\pm \sqrt{x}$ is a better practice to show you know there are 2 square roots. $\endgroup$ – Narasimham Mar 22 '15 at 6:08
2
$\begingroup$

You have hit upon one of the differences between $\sqrt{x}$ and $x^{\frac{1}{2}}$.

In particular, fractional powers are multivalued by nature. $4^{\frac{1}{2}}= \{2,-2\}$, however when using radicals, we are interested only in the principal root ($\sqrt{x}=|x^{\frac{1}{2}}|$), $\sqrt{4} = 2$.

Indeed, if you were to graph $x^{\frac{1}{2}}=y$ it will be the same as the graph for $x=y^2$.

Since $\sqrt{x}$ however outputs exactly one value for each $x$, you have that $y=\sqrt{x}$ is a function (i.e. passes the vertical line test) and will include only the positive arc.


It should be mentioned that the action you describe of "squarerooting" both sides is, although intuitive, technically incorrect. $x=y^2\Rightarrow \sqrt{x}=\sqrt{y^2}\Rightarrow \sqrt{x}=|y|$ which is different than simply $y$. When you use a squareroot to cancel out a power, you must include absolute value signs.

$\endgroup$
1
$\begingroup$

A square root of $a$ is any number $b$ such that $b^2=a$. Thus, for all nonzero numbers, there are exactly two square roots (this is why there are two $y$-values for each positive $x$-value in $x=y^2$). The radical notation $\sqrt{a}$ refers to the principal or positive square root. So both $2$ and $-2$ are square roots of $4$, but $\sqrt{4}$ is the positive square root, which is $2$.

$\endgroup$
  • $\begingroup$ The principal square root is the nonnegative square root. $\endgroup$ – N. F. Taussig Mar 22 '15 at 11:45
  • $\begingroup$ It depends on who you ask. Because $\sqrt{0}$ is not multivalued, the word "positive" or "negative" is used to define what happens to positive numbers (when restricted to the real case). There is never a need to clarify what is occurring at $0$. The same reasoning extends to discussing the complex generalization. There's certainly nothing wrong with "nonnegative," but "positive" is commonly used as well. $\endgroup$ – rnrstopstraffic Mar 22 '15 at 16:50
0
$\begingroup$

Because as you said you cannot square root a negative number, but consider $y^2=x$ if y=-2 you have 4=x, if y=2 you have 4=x

i.e., you can have 2 but also -2.

$\endgroup$
0
$\begingroup$

The graph of $x = y^2$ contains points $(x,y)$ with $y$ negative while the graph of $y = \sqrt{x}$ has points $(x,y)$ with $y \geq 0$. Viewed as sets of points, they do not have the same elements which are ordered pairs ,hence the graphs are different.

$\endgroup$
  • $\begingroup$ Thanks this helped. I just find it weird that they don't explain why something like this can make a difference in school $\endgroup$ – gopro_2027 Mar 22 '15 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.