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

Oct 28, 2017

Solution: $x = 6 + 2 i \mathmr{and} x = 6 - 2 i$

Explanation:

Let ${x}^{2} - 12 x + 40 = 0 \mathmr{and} \left({x}^{2} - 12 x\right) + 40 = 0$ or

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

${\left(x - 6\right)}^{2} = - 4 \mathmr{and} \left(x - 6\right) = \pm \sqrt{- 4}$ or

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

$x - 6 = \pm 2 i \mathmr{and} x = 6 \pm 2 i \therefore x = 6 + 2 i \mathmr{and} x = 6 - 2 i$

Solution: $x = 6 + 2 i \mathmr{and} x = 6 - 2 i$ [Ans]