How do you solve $2 x^{2} - 8 x + 16 = 0$ $2x^2 - 8x + 16 = 0$ by completing the square?

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How do you solve $2 x^{2} - 8 x + 16 = 0$ $2x^2 - 8x + 16 = 0$ by completing the square?

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Answer 1 · 2016-04-10T10:34:53

$x = 2 - 2 i$ $x=2-2i$ or $x = 2 + 2 i$ $x=2+2i$

Explanation:

Each monomial in equation $2 x^{2} - 8 x + 16 = 0$ $2x^2-8x+16=0$ , is divisible by $2$ $2$ , so diving by $2$ $2$ , we get

$x^{2} - 4 x + 8 = 0$ $x^2-4x+8=0$

Now comparing it with ${(x - a)}^{2} = x^{2} - 2 a x + a^{2}$ $(x-a)^2=x^2-2ax+a^2$ , it is apparent that if $- 2 a = - 4$ $-2a=-4$ , we have $a = 2$ $a=2$ and $a^{2} = 4$ $a^2=4$ .

Hence adding $4$ $4$ , we can complete the square. Doing so gives us

$x^{2} - 4 x + 4 - 4 + 8 = 0$ $x^2-4x+4-4+8=0$ or

${(x - 2)}^{2} + 4 = 0$ $(x-2)^2+4=0$ but as in the domain of real numbers the LHS of the equation will be always positive, we cannot have real roots but only complex roots. Hence we write $+ 4$ $+4$ as #-(2i)^2 and then equation becomes

${(x - 2)}^{2} - {(2 i)}^{2} = 0$ $(x-2)^2-(2i)^2=0$ or $(x - 2 + 2 i) (x - 2 - 2 i) = 0$ $(x-2+2i)(x-2-2i)=0$

i.e. $x = 2 - 2 i$ $x=2-2i$ or $x = 2 + 2 i$ $x=2+2i$

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How do you solve $2 x^{2} - 8 x + 16 = 0$ $2x^2 - 8x + 16 = 0$ by completing the square?

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Explanation:

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How do you solve $2 x^{2} - 8 x + 16 = 0$ $2x^2 - 8x + 16 = 0$ by completing the square?