Squaring both sides of an equation?

[bobyo]

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

 

[Haynes Kwon]

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.

 

[PeroK]

What makes you think that would be wrong?

 

[Haynes Kwon]

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.

 

[PeroK]

If I wanted to simplify an equation, say of the form [tex]\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?

Thanks

It depends what you want to do. Suppose, for example, $B = -3$, then your equation becomes:

$\sqrt{A} = -B$

$A = B^2 = 9$

Whereas, squaring the original equation doesn't get you very far:

$A -6 \sqrt{A}+ 9 = 0$

Which doesn't really help.

 

[Mark44]

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".

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.

 

[Haynes Kwon]

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.