Squaring each side of the equation

[Yankel]

Hello all,

I am working on a calculus problem, of finding a local min and max of a function with 2 variables. During my solution, I have encountered an algebraic issue, maybe you could assist.

I am trying to solve the equation:

\[7x-1=\sqrt{56x^{2}-16x-31}\]

If I let MAPLE solve it, I get one solution: x=16/7, which is identical to the solution in the book (one critical point in the calculus view).

What I did from here, is:

\[(7x-1)^{2}=56x^{2}-16x-31\]The solution I got to that, which is what MAPLE gives to that, is two points: x=16/7 and x=-2.

I don't see what I did wrong here, I used the power on both sides, on the entire side, and not on elements, like you should do. Can you help ?

If you are curious, the function is:

\[f(x,y)=\sqrt{56x^{2}-8y^{2}-16x-31}+1-8x\]Many thanks !

 

[kaliprasad]

Hello all,

I am working on a calculus problem, of finding a local min and max of a function with 2 variables. During my solution, I have encountered an algebraic issue, maybe you could assist.

I am trying to solve the equation:

\[7x-1=\sqrt{56x^{2}-16x-31}\]

If I let MAPLE solve it, I get one solution: x=16/7, which is identical to the solution in the book (one critical point in the calculus view).

What I did from here, is:

\[(7x-1)^{2}=56x^{2}-16x-31\]The solution I got to that, which is what MAPLE gives to that, is two points: x=16/7 and x=-2.

I don't see what I did wrong here, I used the power on both sides, on the entire side, and not on elements, like you should do. Can you help ?

If you are curious, the function is:

\[f(x,y)=\sqrt{56x^{2}-8y^{2}-16x-31}+1-8x\]Many thanks !

because you squared both the sides you got one erroneous root x = -2 gives the LHS -ve but RHS is positive and so this is not solution

note that $2^2 = (-2)^2 = 4$ but $\sqrt{4} = 2$

 

[Yankel]

I am not sure I get it, why not:

\[\sqrt{4}=\pm 2\]

?

I understand that if I put x=-2 before squaring, there is no solution.

But how can you know when to square and not to square ?

From solving inequalities with roots, I know that you can square as long as you do it on the entire LHS and the entire RHS.

How would you solve it ?

 

[Dustinsfl]

I am not sure I get it, why not:

\[\sqrt{4}=\pm 2\]

?

I understand that if I put x=-2 before squaring, there is no solution.

But how can you know when to square and not to square ?

From solving inequalities with roots, I know that you can square as long as you do it on the entire LHS and the entire RHS.

How would you solve it ?

For real solution, \(\sqrt{y}\geq 0\). When x =-2, \(7x - 1 = -15 < 0\). So then you have \(-15 = \sqrt{y}\) but how can this be when the sqrt is greater than or equal to 0?

 

[Deveno]

By convention, $\sqrt{a}$ always means the non-negative root.

The reason why we do this is so that we can talk about the function $f$, where:

$f(x) = \sqrt{x}$.

Now the squaring function (the parabola $g(x) = x^2$) is not one-to-one, it obviously takes $-a$ and $a$ to the same image ($a^2$). So when we square, we get two possible ways to "unsquare", so we ALWAYS have to check if the two possibilities we get when we solve by "taking squares" fit the original problem.

One sees this a lot in mathematical modelling of physical problems, where often the "answer" is meant to be some dimensional entity, like length or area. If we get a negative answer, after squaring somewhere, it is typically discarded as "extraneous" (saying something is -5 cm long doesn't make too much sense).

With "abstract problems" involving functions, it is not often clear WHICH solution is "the right one", and both may be viable candidates. This is often true in situations involving angles, where a negative angle might mean "reflection from below" instead of "reflection from above".

With REAL numbers, we have a way of distinguishing one square root as "bigger" (since the real numbers are ORDERED), which gives a clear way of picking a "preferred" square root: the bigger one (the positive one). This turns out to be fantastically UNTRUE in the complex numbers, leading to such amusing things as:

$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i^2 = -1$

(the problem here is two-fold: first $\sqrt{(-1)} = \pm i$ so the next to last statement is suspect, and also $\sqrt{1} = \pm 1$ so the first statement is suspect, we deliberately choose "the wrong square root" at the start to make it seem as if the impossible $1 = -1$ might plausibly be true. Because of this, the "rule":$\sqrt{ab} = \sqrt{a}\sqrt{b}$

is not allowed for complex numbers, except in certain very special circumstances (typically $a,b \in \Bbb R^+_0$), there's too many ways in which it can be abused).

Put another way:

if $A = B$, then certainly $A^2 = B^2$, but if we START with:

$A^2 = B^2$, we do not know if $A = B$, or $A = -B$.