Negative square root in system of equations? sqrt(x)+2*sqrt(y)=3, 2*sqrt(x)+sqrt(y)=0

[Ilikebugs]

sqrt(x)+2*sqrt(y)=3 and 2*sqrt(x)+sqrt(y)=0.

I keep getting sqrt(y)=2 and sqrt(x)=-1, but shouldn't sqrt(x) be positive? I'm assuming that x isn't i^4 or something equivalent.

 

[Dr.Peterson]

sqrt(x)+2*sqrt(y)=3 and 2*sqrt(x)+sqrt(y)=0.

I keep getting sqrt(y)=2 and sqrt(x)=-1, but shouldn't sqrt(x) be positive? I'm assuming that x isn't i^4 or something equivalent.

If the context only allows real values, then you have shown that there is no solution. That is a valid answer.

If the context allows complex numbers, then we have a slightly different issue, because square roots have to be taken as two-valued "functions", and this sort of equation really shouldn't be written. But if you did that, then x = 1^2 = 1 -- you would have a real solution, only by virtue of interpreting the symbol differently. I doubt that this is what you are supposed to do.

Note that i^4 = 1, so that doesn't help.

Might you have copied the problem wrong?

 

[sinx]

sqrt(x)+2*sqrt(y)=3 and 2*sqrt(x)+sqrt(y)=0.

I keep getting sqrt(y)=2 and sqrt(x)=-1, but shouldn't sqrt(x) be positive? I'm assuming that x isn't i^4 or something equivalent.

sq roots can be + or -.
plug your possible solns back into the initial equations.
you will see that -1 is the soln for sqrt x, (and y=4).

what you can't do is take the sqrt of -1, (except in the imaginary sense).

 

[Dr.Peterson]

sq roots can be + or -.
plug your possible solns back into the initial equations.
you will see that -1 is the soln for sqrt x, (and y=4).

what you can't do is take the sqrt of -1, (except in the imaginary sense).

Actually, no. Although any positive number has two square roots, when we use the radical symbol √, or "sqrt()", we are referring to the square root function, which can have only one value; to make it a function, we define it to mean only the principal square root, which is the non-negative one. (When we want both roots, we use the ± symbol to make that explicit, as in the quadratic formula. We do the same when we solve an equation by taking the square root.)

So there is no solution; the OP had it right. If you were to continue your thinking and say that x=1, y=4, you would have this:

sqrt(x)+2*sqrt(y)=3 ==> sqrt(1)+2*sqrt(4)=3 ==> 1+2*2=3 is false
2*sqrt(x)+sqrt(y)=0 ==> 2*sqrt(1)+sqrt(4)=0 ==> 2*1+2=0 is false

My comment was that in some contexts, we do allow the radical to be multiple-valued, representing both roots at once; in the case of complex numbers, there is no principal root, so if we use it at all, it has to be this way. But this is not that context. You don't write equations that way.

 

[pka]

Actually, no. Although any positive number has two square roots, when we use the radical symbol √, or "sqrt()", we are referring to the square root function, which can have only one value; to make it a function, we define it to mean only the principal square root, which is the non-negative one. (When we want both roots, we use the ± symbol to make that explicit, as in the quadratic formula. We do the same when we solve an equation by taking the square root.)

So there is no solution; the OP had it right. If you were to continue your thinking and say that x=1, y=4, you would have this:

sqrt(x)+2*sqrt(y)=3 ==> sqrt(1)+2*sqrt(4)=3 ==> 1+2*2=3 is false
2*sqrt(x)+sqrt(y)=0 ==> 2*sqrt(1)+sqrt(4)=0 ==> 2*1+2=0 is false

My comment was that in some contexts, we do allow the radical to be multiple-valued, representing both roots at once; in the case of complex numbers, there is no principal root, so if we use it at all, it has to be this way. But this is not that context. You don't write equations that way.

@Prof. Peterson, thank you for this correction, it was sorely needed. There is mass confusion on this topic. In the last best practices panel I attended there was consensus that the square root function be restricted to non-negative real numbers.
But as History will do, we were forced to eccept the usage where if \(\displaystyle x<0\) then \(\displaystyle \sqrt{x}=\sqrt{|x|}\bf{i}.\)
But let me be clear, we are working in a model theory setting, an enlargement of the reals. The symbol \(\displaystyle \bf{i}\) is a solution to the real equation \(\displaystyle x^2+1=0\).
To be clear, this enlargement of the reals is the complex field and \(\displaystyle \bf{i}^2=-1\).
In general, if \(\displaystyle z\in\mathbb{C}\), a complex number, then there are \(\displaystyle \bf{n}\) nth roots of \(\displaystyle z\). BUT we never, never use the notation \(\displaystyle \sqrt[n]{z}\).

 

[Dr.Peterson]

@Prof. Peterson, thank you for this correction, it was sorely needed. There is mass confusion on this topic. In the last best practices panel I attended there was consensus that the square root function be restricted to non-negative real numbers.
But as History will do, we were forced to accept the usage where if \(\displaystyle x<0\) then \(\displaystyle \sqrt{x}=\sqrt{|x|}\bf{i}.\)
But let me be clear, we are working in a model theory setting, an enlargement of the reals. The symbol \(\displaystyle \bf{i}\) is a solution to the real equation \(\displaystyle x^2+1=0\).
To be clear, this enlargement of the reals is the complex field and \(\displaystyle \bf{i}^2=-1\).
In general, if \(\displaystyle z\in\mathbb{C}\), a complex number, then there are \(\displaystyle \bf{n}\) nth roots of \(\displaystyle z\). BUT we never, never use the notation \(\displaystyle \sqrt[n]{z}\).

One thing I've observed is that authors find it necessary to define the square root of a negative number as you did, just so they can use the quadratic formula when the discriminant is negative. I tell my students about the problems that arise from this [a single-valued square root does not obey the product rule, that sqrt(a)*sqrt(b) = sqrt(ab), for negative numbers], and say that they must always use ± with such radicals, and must first express in terms of i before doing any simplifying. A dangerous notation requires rules for safe handling!

 

[pka]

One thing I've observed is that authors find it necessary to define the square root of a negative number as you did, just so they can use the quadratic formula when the discriminant is negative. I tell my students about the problems that arise from this [a single-valued square root does not obey the product rule, that sqrt(a)*sqrt(b) = sqrt(ab), for negative numbers], and say that they must always use ± with such radicals, and must first express in terms of i before doing any simplifying. A dangerous notation requires rules for safe handling!

You are of course correct about the quadratic formula being the reason for the concession.
And I should have excluded the product rule. Here is more on that.
\(\displaystyle \left( {\sqrt { - 8} } \right)\left( {\sqrt { - 2} } \right) = \left( {\sqrt 8 i} \right)\left( {\sqrt 2 i} \right) = - 4 \ne \sqrt {16} \)

BUT the is a logical solution.
\(\displaystyle \begin{array}{l} \left( {\sqrt { - 8} } \right)\left( {\sqrt { - 2} } \right)\\ {\left( {8\exp (\pi ~i)} \right)^{\frac{1}{2}}}{\left( {2\exp (\pi~ i)} \right)^{\frac{1}{2}}}\\ \left( {\sqrt 8 \exp \left( {\frac{{\pi i}}{2}} \right)} \right)\left( {\sqrt 2 \exp \left( {\frac{{\pi i}}{2}} \right)} \right)\\ \left( {\sqrt {16} \exp (i\pi )} \right)\\ - 4 \end{array}\)

Although that first step appears to be a violation of notation, it does show how we can proceed to regular order.