Squaring both sides of an equation?

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The discussion centers on the mathematical process of squaring both sides of the equation \(\sqrt{A} + B = 0\) to eliminate the square root. It is established that squaring without transposing terms can lead to the inclusion of linear terms, complicating the equation. The correct approach involves recognizing the constraints on A and B, specifically that A must be non-negative (A ≥ 0) and B must be non-positive (B ≤ 0). The simplification to \(A = B^2\) is valid under these conditions, but squaring the equation directly without rearranging can yield unnecessary complexity.

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If I wanted to simplify an equation, say of the form \sqrt{A} + B = 0 to get rid of the square root, is it correct to square as is? If so, why would it then be wrong to move one term to the other side before squaring?

Thanks
 
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Prior to beginning, you must be aware of the possible ranges of A and B.
I believe this is what you missed.
A>0 or A=0
I will assume that B is the component of the set of the entire negative real number.

If you have learned about how
(X+Y)^2 is developed,

X^2+2XY+Y^2,

you will see that the linear terms of X and Y still remain in the equation.

Now let's substitute A^(1/2) with P, and B with Q. In order to get rid of square root, every degree of P should be the multiple of 2, or 0.

P + Q = 0

If you square both sides of the equation without any transposition, that will be

P^2+2PQ+Q^2=0

and you will notice that still there is a linear term of P.

If you just want to simplify the original equation,

A=B^2 (A>0 or A=0, B<0 or B=0)

may be accurate.
 
Last edited:
What makes you think that would be wrong?
 
B must be 0 or a negative number. If he squared with one term moved to other side, he could have got a positive value.
 
Last edited:
bobyo said:
If I wanted to simplify an equation, say of the form \sqrt{A} + B = 0[/teAx] to get rid of the square root, is it correct to square as is? If so, why would it then be wrong to move one term to the other side before squaring?<br /> <br /> Thanks
<br /> <br /> It depends what you want to do. Suppose, for example, ##B = -3##, then your equation becomes:<br /> <br /> ##\sqrt{A} = -B##<br /> <br /> ##A = B^2 = 9##<br /> <br /> Whereas, squaring the original equation doesn&#039;t get you very far:<br /> <br /> ##A -6 \sqrt{A}+ 9 = 0##<br /> <br /> Which doesn&#039;t really help.
 
Haynes Kwon said:
Prior to beginning, you must be aware of the possible ranges of A and B.
I believe this is what you missed.
A>0 or A=0
I will assume that B is the component of the set of the entire negative real number.

If you have learned about how
(X+Y)^2 is developed,

X^2+2XY+Y^2,

you will see that the linear terms of X and Y still remain in the equation.

Now let''s substitute A^(1/2) with P, and B with Q. In order to get rid of square root, every degree of P should be the multiple of 2, or 0.

P + Q = 0

If you square the equation without any transposition, that will be

P^2+2PQ+Q^2
If you "square an equation" (really, square both sides of an equation), you should end up with an equation.
What you have above is missing "= 0".
Haynes Kwon said:
and you will notice that still there is a linear term of P.

If you just want to simplify the original equation,

A=B^2 (A>0 or A=0, B<0 or B=0)

may be accurate.
 
Mark44 said:
If you "square an equation" (really, square both sides of an equation), you should end up with an equation.
What you have above is missing "= 0".

Thank you. My bad.
 

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