1
$\begingroup$

Is it wrong to say that 100 is solution of $\sqrt x +10=0$?

I know that range of $\sqrt{x}$ is $[0, \infty)$ by convention. This convention is because of definition of a function. So if I consider $\sqrt x+10=0$ as a simple equation then can I say hundred is a solution to the equation?

I think I can say that because if I square both sides I get $x=100$ and also if I consider $\sqrt x+10=0$ just as a equation then there is no need of thinking about functions and the convention of omitting negative roots.

$\endgroup$
  • $\begingroup$ You can say it because $\sqrt{100} - 10 = 0$. So 100 is a solution. That's all there is to it. Why wouldn't you be able to say that. $\endgroup$ – fleablood Apr 20 '17 at 0:35
  • 1
    $\begingroup$ Please see: math.stackexchange.com/questions/809424/… $\endgroup$ – HEKTO Apr 20 '17 at 0:46
  • 2
    $\begingroup$ It wasn't me who downvoted, but I don't disagree with the downvote. Some of the suggested reasons for downvoting: "The question does not show any research effort; it is not clear or not useful". The shortest amount of research into the question of your own will tell you that $\sqrt{x}$ is always positive or zero, and that a positive number plus another positive number is always positive, implying that $\sqrt{100}+10$ is certainly not zero. We only consider the positive root in any context that the $\sqrt{~}$ symbol is used, regardless of whether we are calling it a function or not. $\endgroup$ – JMoravitz Apr 20 '17 at 0:50
  • 2
    $\begingroup$ Ah.. But $\sqrt{n}$ is NOT the negative square root. It is, by DEFINITION, the *positive square root. $\sqrt{n} = -10$ does NOT mean $(-10)^2 = n$. It means $(-10)^2 = $ AND $-10 \ge 0$. Which is simply not true. $\endgroup$ – fleablood Apr 20 '17 at 0:56
  • 1
    $\begingroup$ Yes you do. <><> $\endgroup$ – fleablood Apr 20 '17 at 1:01
3
$\begingroup$

The solutions to $f(x) = k$ are a SUBSET of the solutions to $f(x)^2 = k^2$ but not all the solution to $f(x)^2 = k^2$ are solutions to $f(x) = k$. Squaring both sides of an equation add extraneous solutions.

$\sqrt{x} + 10 = 0$

$\sqrt{x} = -10$ Doesn't just mean that $(\sqrt{x}^2 = (-10)^2$. It ALSO means that $\sqrt{x} = -10 < 0$.

When we square both sides we LOSE information.

$\sqrt{x}^2 = (-10)^2$

$x = 100$ but we have completely LOST that $\sqrt{x} < 0$.

Any $\sqrt{100} + 10 = 10 + 10 = 20 \ne 0$. SO it simply DOESN'T work.

Consider this:

$x =2 $ has one solution. Square both sides and you get $x^2 = 4$. Which has TWO solutions!!! Where did that solution $x = -2$ come from?

It came because when we squared both sides we added invalid extraneous solutions.

Solutions to $x= 2$ is $\{2\}$. Solutions to $x^2 = 4$ has solutions $\{2,-2\}$ and $\{2\} \subset \{2,-2\}$. But it doesn't go the other way. It only goes one way.

$\endgroup$
1
$\begingroup$

Consider the equation x^2 = 100

When we take the square root of the equation to solve it, we would write

x = ± sqrt(100)

Notice that there is a ± sign in front of the square root. The reason that is there is precisely because sqrt(x) is defined as the principle square root.

So, when you evaluate sqrt(100) + 10, you get 10 + 10 = 20, proving your equation false.

$\endgroup$
0
$\begingroup$

$x=100$ is most definitely a solution to $\sqrt{x}-10=0$, because if you plug $x=100$ into the equation, you get $\sqrt{100}-10=0$, or $10-10=0$, which is true.

$\endgroup$
  • $\begingroup$ I think the OP changed the + sign to the minus sign, which he must have meant all along. $\endgroup$ – Mark Fischler Apr 20 '17 at 0:35
  • $\begingroup$ @MarkFischler edited accordingly $\endgroup$ – ASKASK Apr 20 '17 at 0:35
  • 2
    $\begingroup$ @MarkFischler No that was not my edit. $\endgroup$ – A---B Apr 20 '17 at 0:36
-1
$\begingroup$

$√(x)+10=0$

$√(x)=-10$

$√(x)^2=(-10)^2$

$x=100$

$√(100)+10≠0$

Thus, since $√(100)+10$ does not equal zero, this equation has no solution.

$\endgroup$

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.