How do you solve using completing the square method $y = x^{2} - 12 x + 40$ $y=x^2-12x +40$ ?

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Answer 1 · 2017-10-28T10:52:21

Solution: $x = 6 + 2 i and x = 6 - 2 i$ $x= 6+2i and x= 6-2i$

Let $x^{2} - 12 x + 40 = 0 or (x^{2} - 12 x) + 40 = 0$ $x^2-12x+40 =0 or(x^2-12x)+40=0$ or

$(x^{2} - 12 x + 36) - 36 + 40 = 0$ $(x^2-12x +36)-36+40=0$ or

${(x - 6)}^{2} = - 4 or (x - 6) = \pm \sqrt{- 4}$ $(x-6)^2=-4 or (x-6)=+-sqrt (-4)$ or

$(x - 6) = \pm \sqrt{4 i^{2}} [i^{2} = - 1]$ $(x-6)=+-sqrt (4i^2) [i^2=-1]$ or

$x-6 =+-2i or x = 6+- 2i :. x= 6+2i and x= 6-2i$

Solution: $x= 6+2i and x= 6-2i$ [Ans]

How do you solve using completing the square method y=x^2-12x +40y=x2−12x+40y=x^2-12x +40?