How do you solve using the completing the square method $x^{2} - 6 x + 10 = 0$ $x^2 - 6x + 10 = 0$ ?

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Answer 1 · 2017-01-31T04:48:43

$x = 3 \pm i$ $x=3+-i$

$x^{2} - 6 x + 10 = 0$ $x^2-6x+10=0$

consider #coefficient of x, divide by 2

${(x - \frac{6}{2})}^{2} - {(- \frac{6}{2})}^{2} + 10 = 0$ $(x-6/2)^2 -(-6/2)^2+10=0$

${(x - 3)}^{2} - 9 + 10 = 0$ $(x-3)^2-9+10=0$

${(x - 3)}^{2} + 1 = 0$ $(x-3)^2+1=0$

${(x - 3)}^{2} = - 1$ $(x-3)^2=-1$

$i^{2} = - 1$ $i^2=-1$ , then $i = \pm \sqrt{- 1}$ $i=+-sqrt(-1)$

$x - 3 = \pm \sqrt{- 1} = \pm i$ $x-3 = +-sqrt(-1)=+-i$

$x = 3 \pm i$ $x=3+-i$

How do you solve using the completing the square method x^2 - 6x + 10 = 0x2−6x+10=0x^2 - 6x + 10 = 0?