Complete the square, doesn't make sense, QUESTION(s) explain/show?

Question

x²+2ax=b
work:
(x+a)²-2a=b

This would make sense, we would have to complete the square by mutiplying "2" and then squaring it.
(x+a)²-4a²=b
(x+a)²=4a²+b
x+a=√4a²+b
x+a=2a√b
then..
x=a√b

How would it make sense skiping by mutiplying "2", and replacing it by ² to get -a² instead of just 4a²?
(x+a)²-a²=b

see?

Answer 1

x^2 + 2ax = b
The bit to add on to each side is the square of half
the constant that is multiplied by x.
The constant that is multiplied by x is 2a.
Half of 2a is a.
The square of a is a^2.
Thus, x^2 + 2ax + a^2 = b + a^2
or, (x + a)^2 = b + a^2
Therefore, x + a = ± sqrt(b + a^2)
so, x = ± sqrt(b + a^2) - a

Answer 2

Always hated completing the square, and always hated teaching it. If an equation set to zero will not factor, I prefer the quadratic formuna x = -b + or - b^2 etc. google it. If it can't be factored, you can always get the answer this way. Likely you can also google math + Completing the square to get a step by step operation on it. The quadratic formula comes from completing the square of ax^2 + bx +c=O, where a,b and c are real numbers.

Answer 3

x²+2ax=b
work:
(x+a)²-2a=b UM… ERROR HERE, IT’S -2ax NOT -2a

This would make sense, we would have to complete the square by mutiplying "2" and then squaring it.
Multiplying what by 2? You have to multiply EVERYTHING by 2 if you’re going to multiply anything by 2.
You multiplied the second term on the left by 2a. You have to multiply the first tern by 2a and the b on the right side by 2a. You just can’t introduce random terms. The distributive property of multiplication over addition says a(b+c)=ab+ac
(x+a)²-4a²=b THIS STEP IS ABSOLUTELY INADMISSABLE IN THIS CONTEXT If it were true, then from the last erroneous equation, (x+a)²-2a = (x+a)²-4a², or -2a=-4a², or a=1/2 ALWAYS.
(x+a)²=4a²+b
x+a=√4a²+b
x+a=2a√b HOW DO YOU GET FROM HERE
then..
x=a√b TO HERE? If x+a = 2a√b, then you add –a to both sides and get x = 2a√b – a. At best that’s a(-1 + 2√b)

How would it make sense skiping by mutiplying "2", and replacing it by ² to get -a² instead of just 4a²?
(x+a)²-a²=b

see? Nope... don't see...

Completing the Square... here's how.

This is an algebra exercise designed to help solve quadratic equations of the form
ax²+bx+c=0

The idea is to get something that looks like
(dx+e)²=f and, taking the square root of both sides, solve for x.
Multiplying (dx+e)² out, we get (dx)² + 2dex + e²

1. Multiply through 4a
4a²x² + 4abx + 4ac = 0
(2ax)² + 2(2a)bx + 4ac = 0 see where we’re going?

2. Add -4ac to both sides
(2ax)² + 2(2a)bx = -4ac

3. Now, to complete the square on the left, add b², but you have to do it to both sides of the equation.
(2ax)² + 2(2a)bx + b² = b² - 4ac
At this point, we have a perfect square on the left side of the equation…
(2ax + b)² = b² - 4ac

4. Taking the square root of both sides, taking care to account for the fact that a²=(-a)²
2ax + b = ±√(b² - 4ac)

5. The point now is to get x on one side of the equation all by itself. So the next step is to add –b to both sides
2ax = -b ±√(b² - 4ac)

6. The final step, multiply both sides by 1/2a
x = [-b ±√(b² - 4ac)]/(2a)
This is the famous… or infamous… Quadratic Formula.

Another approach is:
ax²+bx+c=0

1. Multiply through by a
a²x²+abx+ac=0

2. Add –ac to both sides
a²x²+abx= -ac

3. Now, add (b/2)² to both sides
a²x²+abx+(b/2)²= b²/4 – ac=(b²-4ac)/4
Now you have
(ax + b/2)²= (b²-4ac)/4
4. Taking the square root of both sides…
ax+b/2 = ±√[(b²-4ac)/4] = ±[√(b²-4ac)]/2

5. Add –b/2 to both sides…
ax = [-b ±√(b²-4ac)]/2

6. Finally, multiply both sides by 1/a
x = [-b ±√(b²-4ac)]/(2a)
Look familiar?

Answer 4

First, get the equation into the form:
x²+ax+b=0
Next complete the square by adding or subtracting the quantity that will make b equal to the square of a/2. This quantity equals (a²/4 - b)

The equation should now be:
x²+ax+(a²/4) = (a²/4 - b)
Factor:
(x+a/2)² = (a²/4 - b)

And solve from there.

Answer 5

The second equation you present is wrong. If
x^2+2ax = b
Then x^2 +2ax +a^2 - a^2 = b and
(x+a)^2 = b+a^2
etc.
In your presentation, you just can flout math rules.