How do you solve $x^{2} + \frac{1}{2} x = 1$ $x^2 + 1/2x = 1$ by completing the square?

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Answer 1 · 2016-07-12T22:23:28

$x = 0.781 or x = - 1.281$ $x = 0.781 " or " x = -1.281$

The constant (1) is already on the RHS.

We need to add in a term so that the left hand side can be factored as ${(x \pm ?)}^{2}$ $(x +-?)^2$

The missing term is found by dividing the coefficient of x by 2 and then squaring the answer. The result must be added to BOTH sides.

$x + \frac{1}{2} x + {(\frac{1}{4})}^{2} = 1 + {(\frac{1}{4})}^{2}$ $x +1/2x + color(red)((1/4))^2 = 1 + color(red)((1/4))^2$

${(x + \frac{1}{4})}^{2} = 1 \frac{1}{16} = \frac{17}{16}$ $(x+1/4)^2 = 1 1/16 = 17/16$

$x + \frac{1}{4} = \pm \sqrt{\frac{17}{16}}$ $x + 1/4 = +-sqrt(17/16)$

$x = \pm \frac{\sqrt{17}}{4} - \frac{1}{4} = \frac{\pm \sqrt{17} - 1}{4}$ $x = +-(sqrt17)/4 -1/4 = (+-sqrt17 -1)/4$

$x = 0.781 or x = - 1.281$ $x = 0.781 " or " x = -1.281$

How do you solve x2+12x=1x^2 + 1/2x = 1 by completing the square?