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readspec1 · Mar 13, 2012, 04:34 PM

solve x^2 x-6=0 by completing the square?

ebaines · Mar 14, 2012, 06:10 AM

To complete the square take half of the 'x' coefficient, square it, and add to both sides.

$ x^2 + x -6 = 0 \\ x^2 + x + (\frac 1 2 )^2 -6 = (\frac 1 2 )^2\\ (x + \frac 1 2 )^2 = (1/2)^2+ 6 = \frac {25} 4\\ x+ \frac 1 2 = \pm \sqrt{\frac {25} 4} =\pm \frac 5 2\\ x = \frac { \pm 5 - 1 } 2 \\ x = 2 \ or \ -3 $

You can compare this to the results using the quadratic equation:

$ x = \frac {-b \pm \sqrt{ b^2 - 4ac}} {2a} = \frac {-1 \pm \sqrt{1-4(1)(-6)}} 2 = \frac {-1 \pm 5} 2 $