How do you solve the quadratic $2 x^{2} - 6 x + 5 = 0$ $2x^2-6x+5=0$ with complex numbers?

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How do you solve the quadratic $2 x^{2} - 6 x + 5 = 0$ $2x^2-6x+5=0$ with complex numbers?

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Answer 1 · 2017-12-15T23:19:15

$x = \frac{3}{2} \pm \frac{1}{2} i$ $x = 3/2+-1/2i$

Explanation:

$0 = 2 (2 x^{2} - 6 x + 5)$ $0 = 2(2x^2-6x+5)$

$0 = 4 x^{2} - 12 x + 10$ $color(white)(0) = 4x^2-12x+10$

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

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

$0 = {(2 x - 3)}^{2} - i^{2}$ $color(white)(0) = (2x-3)^2-i^2$

$0 = ((2 x - 3) - i) ((2 x - 3) + i)$ $color(white)(0) = ((2x-3)-i)((2x-3)+i)$

$0 = (2 x - 3 - i) (2 x - 3 + i)$ $color(white)(0) = (2x-3-i)(2x-3+i)$

So:

$2 x = 3 \pm i$ $2x = 3+-i$

and:

$x = \frac{3}{2} \pm \frac{1}{2} i$ $x = 3/2+-1/2i$

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How do you solve the quadratic $2 x^{2} - 6 x + 5 = 0$ $2x^2-6x+5=0$ with complex numbers?

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How do you solve the quadratic $2 x^{2} - 6 x + 5 = 0$ $2x^2-6x+5=0$ with complex numbers?