Squaring both sides - valid?

**Timbuk2**

If I have an equation, e.g. 2x = 3 + 4y, and I square both sides, does the equation still hold? Is this a valid operation in mathematics?

E.g. If I have 4 > 3 and square both sides, it still holds true. But if you have 3 > -4, and square both sides, it becomes false (9 isn't > than 16).

What about if you have an equation like this?
[latex]|4x + 4| = 5[/latex]

Is squaring both sides an acceptable method for solving it?

Also, if you have an equation like [latex] x = \sqrt{x + 7}[/latex], if you square both sides, you will create an extra solution.

So my question is - is squaring both sides a valid operation in maths?

LeixlipRed

The counter example you gave is missing something vital, it's not an equation, it's an inequality. Squaring both sides is a valid operation when you have an equals sign there. In the last example when you square the restriction on x to be greater than or equal to -7 remains so one solution to the resulting quadratic will only have one valid solution.

LeixlipRed

Actually, that's not quite correct. The domain of the function on the right is {x>=7} but letting x equals to that means that x must be greater than or equal to 0 as the square root must produce a non-negative number. So that restricts solutions to x>=0.

ray giraffe

It depends on the symbol you have in the middle. Each different symbol will need to be investigated separately.

If you have an equals sign then you have the same number on the L.H.S. and R.H.S. (left hand side, right hand side) so you will still have the same number when you square them both.

So the statement x=y implies x^2=y^2, but not the other way around (unless for example we are told and y are positive real numbers). We see that squaring a true equation produces a true equation, but the two equations may not be equivalent (equivalent means each implies the other, if statements are equivalent we think of them as being the "same").

[The function f:x-> x^2 is not injective on the real numbers so f(x)=f(y) does not imply x=y. Applying an injective function to both sides of an equation would produce an equivalent statement.]

Therefore the solutions to L.H.S. = R.H.S. are a subset of the solutions of (L.H.S.)^2 = (R.H.S.)^2, so if we find solutions to the second equation they may not all be solutions to the first equation, they need to be checked.

bnt

**Timbuk2** wrote: »

What about if you have an equation like this?
[latex]|4x + 4| = 5[/latex]

Is squaring both sides an acceptable method for solving it?

If you draw the function graph, you see that this first order equation has two valid solutions, since it's valid when the RHS is -5 as well as +5. So squaring is a good idea here, since it takes the Absolute function out of the equation; just remember that the square root then leads to two solutions.