Why does this even work? (quadratic)

[gavroche]

ax² + bx = a² - b²
a(x² + bx/a) = a² - b²
a(x² + bx/a + b²/(4a²)) = a² - b² + b²/(4a²)
a(x + b/(2a)²) + b² = a² + b²/(4a²)
sqrroot(a) (x + b/(2a)) + b = a + b/(2a)
sqrroot(a) (x + b/(2a)) + b = (2a² + b)/(2a)
a(x + b/(2a)) + ab = (2a³ + ab)/(2a)
(x + b/(2a)) + b = (2a³ + ab)/(2a²)
(x + b/(2a)) + b = a + (ab)/(2a²)
x + b/(2a) + b = a + b/(2a)
x + b = a
x = a - b

So, according to my book, the answer at which I arrived is correct. But I'm a little surprised.
The line that is bothering me is:
a(x² + bx/a + b²/4a²) = a² - b² + b²/4a²

I added b²/4a² to the left side. Shouldn't I have added ab²/4a² to the right side(aka multiply by that a that is outside the paranthesis on the left side) instead of just adding the same b²/4a^{2 ?}

 

[Steven G]

ax² + bx = a² - b²
a(x² + bx/a) = a² - b²
a(x² + bx/a + b²/4a²) = a² - b² + b²/4a²-->you added b²/4a² to the right side but you added a(b²/4a²⁾ to the left side
a(x + b/2a)² + b² = a² + b²/4a² correct, that is it does follow from the line above (which is wrong)
sqrroot(a) (x + b/2a) + b = a + b/2a NO! sqrt(r^2 + s^2 + t^2) is not r + s + t. (sqrt(r^2 * s^2 * t^2) = rst
sqrroot(a) (x + b/2a) + b = (2a² + b)/2a
a(x + b/2a) + ab = (2a³ + ab) / 2a a*sqrt(a) is not sqrt(a). Rather sqrt(a)*sqrt(a)=a
(x + b/2a) + b = (2a³ + ab)/2a²
(x + b/2a) + b = a + (ab)/2a²
x + b/2a + b = a + b/2a
x + b = a
x = a - b

So, according to my book, the answer at which I arrived is correct. But I'm a little surprised.
The line that is bothering me is:
a(x² + bx/a + b²/4a²) = a² - b² + b²/4a²

I added b²/4a² to the left side. Shouldn't I have added ab²/4a² to the right side(aka multiply by that a that is outside the paranthesis on the left side) instead of just adding the same b²/4a^{2 ?} Yes, of course

Ouch, I really do not know what to say. Just read my comments above.
I think that you should set this equation to 0 and then maybe use the quadratic formula.

 

[gavroche]

sqrroot(a) (x + b/2a) + b = a + b/2a NO! sqrt(r^2 + s^2 + t^2) is not r + s + t. (sqrt(r^2 * s^2 * t^2) = rst


I see my error there. However, if I take the square root of: a(x + b/(2a))^2 + b^2 What will the result be then? Since the paranthesis is squared, b is squared, but a is not squared?

ON A SECOND THOUGHT: Taking a square root of a line like that, is probably just a bad idea. The entire equation should be rearranged, right?

 

[Steven G]

Consider the following

3^2 +4^2 = 9 + 16 = 25
So sqrt (3^2 +4^2) = sqrt(9 + 16) = sqrt(25) = 5
You think that sqrt (3^2 +4^2) = 3+4 = 7
We do not get the same answer as 5 does not equal 7.
Your way is wrong!

 

[gavroche]

Ok so, if I had an equation with one side equal to: sqrroot(x^2 + y^2)

The only way to get rid of that square root would be to multiply the entire equation by sqrroot(x^2 + y^2) ?

 

[Steven G]

Ok so, if I had an equation with one side equal to: sqrroot(x^2 + y^2)

The only way to get rid of that square root would be to multiply the entire equation by sqrroot(x^2 + y^2) ?

I would not say the only way as you could multiply sqrt(x^2 + y^2) by 14*sqrt(x^2 + y^2) or by (ab^2)sqrt(x^2 + y^2) or (x^2 + y^2)^1.5 or ... and the sqrt sign will no longer be there.

 

[Dale10101]

Perhaps

ax² + bx = a² - b²
a(x² + bx/a) = a² - b²
a(x² + bx/a + b²/(4a²)) = a² - b² + b²/(4a²)
a(x + b/(2a)²) + b² = a² + b²/(4a²)
sqrroot(a) (x + b/(2a)) + b = a + b/(2a)
sqrroot(a) (x + b/(2a)) + b = (2a² + b)/(2a)
a(x + b/(2a)) + ab = (2a³ + ab)/(2a)
(x + b/(2a)) + b = (2a³ + ab)/(2a²)
(x + b/(2a)) + b = a + (ab)/(2a²)
x + b/(2a) + b = a + b/(2a)
x + b = a
x = a - b

So, according to my book, the answer at which I arrived is correct. But I'm a little surprised.
The line that is bothering me is:
a(x² + bx/a + b²/4a²) = a² - b² + b²/4a²

I added b²/4a² to the left side. Shouldn't I have added ab²/4a² to the right side(aka multiply by that a that is outside the paranthesis on the left side) instead of just adding the same b²/4a^{2 ?}

Perhaps the simplest way to solve for x is to let c = a²-b², then use the quadratic formula to solve ax² + bx = c and then sub a²-b² back into the result. The solution is a good deal more ugly then x = (x-a) or x = (x-a).

When using the perfect square method the term you are adding to each side is a(b/2a)², i.e.

a(x²+bx/a) + a(b/2a)² = a² - b2​ + a(b/2a)²