Solve the quadratic equation by completing the square x^2-12+7=0

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Solve the quadratic equation by completing the square x^2-12+7=0

Mathematics
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  • phi
if you have time, this might help http://www.khanacademy.org/math/algebra/quadtratics/v/completing-the-square-2
Humm. Quadratic... factor? complete the square? or quadratic formula?

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Ah ha!! instructions say "completing the square", so I guess we'll completing the square :)
BTW... was the problem x^2-12x+7=0 (original was missing the x on the 12)
oops yes:)
cool. x^2 -12x + 7 = 0 step 1: move the constant over x^2-12x= -7
step 2: take half the bx term \(12 \div 2 = 6\)
step 3: square the number in step # \(6^2 = 36\)
step 4: add the number from step 3 to both sides: \[x^2 + 12x + 36 = -7 + 36\]
so far so good?
Yep:)
\[x^2 + 12x + 7 = 0\] \[x^2 + 12x = -7\] \[x^2 + 12x + 36 = -7+36\] 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
from this point, use the square root property to finish solving for x.
x^2+12x+7=0?
that is what I started with?
did you just FOIL (x+6)^2 and subtract 29 to zero out the equation! Yep, that is back to where you started... :)
"Square Root Property" means take the square root \(\sqrt{\;\;\;}\) of both sides .
\[\sqrt{(x+6)^2} = \pm \sqrt {29}\]
Ohh!!!:D
:)
I got x=+- -.6148
\[\sqrt{(x+6)^2} = \pm \sqrt {29}\] 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 \[x = -6 - \sqrt{29} \approx -11.385 \]
Thank you sooo much!!!!!:D
np :)

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