How do you solve $x^{2} - 3 x + 6 = 0$ $x^2 - 3x + 6 = 0$ by completing the square?

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Answer 1 · 2017-07-05T07:58:27

$x = \frac{3}{2} \pm i \frac{\sqrt{15}}{2}$ $x=3/2+-isqrt15/2$

$x^{2} - 3 x + 6 = 0$ $x^2-3x+6=0$

add and subtract half the coefficient of $x$ $x$ squared

$(x^{2} - 3 x + {(\frac{3}{2})}^{2}) + 6 - {(\frac{3}{2})}^{2} = 0$ $color(blue)((x^2-3x+(3/2)^2))+6- ( 3/2)^2=0$

the section in blue is a perfect square

${(x - \frac{3}{2})}^{2} + \frac{15}{4} = 0$ $color(blue)((x-3/2)^2)+15/4=0$

solve by rearranging for $x$ $x$

${(x - \frac{3}{2})}^{2} = - \frac{15}{4}$ $(x-3/2)^2=-15/4$

$x - \frac{3}{2} = \sqrt{- \frac{15}{4}}$ $x-3/2=sqrt(-15/4)$

$x - \frac{3}{2} = \pm i \frac{\sqrt{15}}{2}$ $x-3/2=+-isqrt15/2$

$x = \frac{3}{2} \pm i \frac{\sqrt{15}}{2}$ $x=3/2+-isqrt15/2$

How do you solve x2−3x+6=0x^2 - 3x + 6 = 0 by completing the square?