anonymous
  • anonymous
x^2+12x+6=0 complete the square
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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zzr0ck3r
  • zzr0ck3r
x^2+12x=-6 (x+6)^2=-6+36 (x+6)^2-30=y
anonymous
  • anonymous
\[ax^{2} + bx + c = 0\\a \left[ x ^{2} + \left( \frac{ b }{ a } \right)x \right] = -c\\a \left[ x ^{2} + \left( \frac{ b }{ a } \right)x +\frac{ b ^{2} }{ 4a ^{2} } \right] = \frac{ b ^{2} }{ 4a } -c\\x ^{2} + \left( \frac{ b }{ a } \right)x +\frac{ b ^{2} }{ 4a ^{2} } = \frac{ b ^{2} }{ 4a ^{2} } -\frac{ c }{ a }\\\left( x + \frac{ b }{ 2a } \right)^{2} = \frac{ b ^{2} - 4ac }{ 4a ^{2} }\\x + \frac{ b }{ 2a } = \frac{ \pm \sqrt{b ^{2} - 4ac } }{ 2a }\\x = \frac{ -b \pm \sqrt{b ^{2} - 4ac } }{ 2a }\]
zzr0ck3r
  • zzr0ck3r
im not sure the derivation of the quadratic equation will help here....

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anonymous
  • anonymous
I'm sure it would.
zzr0ck3r
  • zzr0ck3r
I think if one cant complete the square then that looks like chinese.
anonymous
  • anonymous
That IS how one goes about doing it.
anonymous
  • anonymous
Completing the square and the quadratic formula, though different approaches, are essentially the same idea.
zzr0ck3r
  • zzr0ck3r
I know...Ithink about the first time you learned to complete the square. If someone asked you to do it, and then showed you that derivation, do you think you could do it?
zzr0ck3r
  • zzr0ck3r
just my opinion, not a big point:0
anonymous
  • anonymous
Think back? Too long! Over 40 years ago! I can't remember what I had for breakfast yesterday! lol ! The common theme is getting the initial x^2 and x terms together so that they can be included under the same square. They both complete the square.
zzr0ck3r
  • zzr0ck3r
lol touche
anonymous
  • anonymous
The thing is, for equations of the form: ax^2 + bx + c = 0 completing the square and the quadratic formula amount to the same thing. For equations of the form: ax^2 + bx + c = y That's a little different, like finding vertex forms. At that point, completing the square diverges from the quadratic formula.

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