Question

Did I "derive" this quadratic equation right?

Answer 1

\[ax^2+bx+c=0\]\[x^2+\frac{ b }{ a }x+\frac{ c }{ a }=0\]\[x^2+\frac{ b }{ a }x=-\frac{ c }{ a }\]\[x^2+\frac{ b }{ a }x+(\frac{ b }{ 2a })^2=-\frac{ c }{ a }+\frac{ b }{ 2a }\]\[(x \frac{ b }{ 2a })^2=-\frac{ c }{ a }+\frac{ b }{ 2a }\]\[x+\frac{ b }{ 2a }=\sqrt{-\frac{ c }{ a }+)\frac{ b }{ 2a })^2}\]\[x=-\frac{ b }{ 2a }\pm \sqrt{-\frac{ c }{ a }+(\frac{ b }{ 2a })^2}=-\frac{ b }{ 2a }\pm \sqrt{\frac{ b^2-4ac }{ 4a^2 }}=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 2

Umm I see an error

When you were supposed to add \[\large (\frac{b}{2a})^2\] to both sides

Answer 3

After line 4, right? >.>

Answer 4

Oh I see what I did there!

Answer 5

I mean granted you accounted for it later on (it just reappeared) wouldn't want credit taken away for missing it

Answer 6

Now that look back in my (messier) scrap paper I see I added \[\frac{ b }{ 2a }\]instead of the actual \[(\frac{ b }{ 2a })^2\]

Answer 7

Yeah...just watch that...but like I said it kind of..."magically reappeared" under your radical 2 lines later

Answer 8

Both other than that and missing a '+' sign in the 5th line

\[\large (x \color \red{+}\frac{ b }{ 2a })^2=-\frac{ c }{ a }+\frac{ b }{ 2a }\]

Everything looks good!

Answer 9

Oh okay, thanks! Phew! :D

Answer 10

No problem!