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

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

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Answer 1 · 2016-04-10T10:34:53

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

Explanation:

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

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

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

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

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

$(x-2)^2+4=0$ 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 $+4$ as #-(2i)^2 and then equation becomes

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

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

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

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