Question

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

Answer 1

@cruffo

Answer 2

if you have time, this might help
http://www.khanacademy.org/math/algebra/quadtratics/v/completing-the-square-2

Answer 3

Humm. Quadratic... factor? complete the square? or quadratic formula?

Answer 4

Ah ha!! instructions say "completing the square", so I guess we'll completing the square :)

Answer 5

BTW... was the problem x^2-12x+7=0 (original was missing the x on the 12)

Answer 6

oops yes:)

Answer 7

cool.
x^2 -12x + 7 = 0
step 1: move the constant over
x^2-12x= -7

Answer 8

step 2: take half the bx term
\(12 \div 2 = 6\)

Answer 9

step 3: square the number in step #
\(6^2 = 36\)

Answer 10

step 4: add the number from step 3 to both sides:
\[x^2 + 12x + 36 = -7 + 36\]

Answer 11

so far so good?

Answer 12

Yep:)

Answer 13

\[x^2 + 12x + 7 = 0\]
\[x^2 + 12x = -7\]
\[x^2 + 12x + 36 = -7+36\]
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

Answer 14

from this point, use the square root property to finish solving for x.

Answer 15

x^2+12x+7=0?

Answer 16

that is what I started with?

Answer 17

did you just FOIL (x+6)^2 and subtract 29 to zero out the equation! Yep, that is back to where you started... :)

Answer 18

"Square Root Property" means take the square root \(\sqrt{\;\;\;}\) of both sides .

Answer 19

\[\sqrt{(x+6)^2} = \pm \sqrt {29}\]

Answer 20

Ohh!!!:D

Answer 21

:)

Answer 22

I got x=+- -.6148

Answer 23

\[\sqrt{(x+6)^2} = \pm \sqrt {29}\]
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
\[x = -6 - \sqrt{29} \approx -11.385 \]

Answer 24

Thank you sooo much!!!!!:D

Answer 25

np :)