anonymous
  • anonymous
Solve the quadratic equation by completing the square x^2-12+7=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@cruffo
phi
  • phi
if you have time, this might help http://www.khanacademy.org/math/algebra/quadtratics/v/completing-the-square-2
cruffo
  • cruffo
Humm. Quadratic... factor? complete the square? or quadratic formula?

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cruffo
  • cruffo
Ah ha!! instructions say "completing the square", so I guess we'll completing the square :)
cruffo
  • cruffo
BTW... was the problem x^2-12x+7=0 (original was missing the x on the 12)
anonymous
  • anonymous
oops yes:)
cruffo
  • cruffo
cool. x^2 -12x + 7 = 0 step 1: move the constant over x^2-12x= -7
cruffo
  • cruffo
step 2: take half the bx term \(12 \div 2 = 6\)
cruffo
  • cruffo
step 3: square the number in step # \(6^2 = 36\)
cruffo
  • cruffo
step 4: add the number from step 3 to both sides: \[x^2 + 12x + 36 = -7 + 36\]
cruffo
  • cruffo
so far so good?
anonymous
  • anonymous
Yep:)
cruffo
  • cruffo
\[x^2 + 12x + 7 = 0\] \[x^2 + 12x = -7\] \[x^2 + 12x + 36 = -7+36\] step 5: factor the left-hand side, and simplify the right-hand side: \[(x+6)^2 = 29\] Hint - the left-hand side always factors to (x + # from step 2, including the sign, either + or -)^2
cruffo
  • cruffo
from this point, use the square root property to finish solving for x.
anonymous
  • anonymous
x^2+12x+7=0?
anonymous
  • anonymous
that is what I started with?
cruffo
  • cruffo
did you just FOIL (x+6)^2 and subtract 29 to zero out the equation! Yep, that is back to where you started... :)
cruffo
  • cruffo
"Square Root Property" means take the square root \(\sqrt{\;\;\;}\) of both sides .
cruffo
  • cruffo
\[\sqrt{(x+6)^2} = \pm \sqrt {29}\]
anonymous
  • anonymous
Ohh!!!:D
cruffo
  • cruffo
:)
anonymous
  • anonymous
I got x=+- -.6148
cruffo
  • cruffo
\[\sqrt{(x+6)^2} = \pm \sqrt {29}\] humm.. \(\sqrt{29}\) does not simplify (other than decimal approx) so I'm gonna leave it for now... \[x+6 = \pm \sqrt{29}\] subtract 6 from both sides (but not from 29, that's inside the square root) \[x = -6 \pm \sqrt{29}\] that gives us two solutions: \[x = -6 + \sqrt{29} \approx -0.615\] and \[x = -6 - \sqrt{29} \approx -11.385 \]
anonymous
  • anonymous
Thank you sooo much!!!!!:D
cruffo
  • cruffo
np :)

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