How do you solve $x^2+8x+1=0$ by completing the square?

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Answer 1 · 2016-06-06T08:51:56

$x=-4-sqrt15$ or

$x=-4+sqrt15$

In $x^2+8x+1=0$ , we observe that coefficient of $x$ is $8$ and hence, we can complete the square by adding square of half of $8$ i.e. $4^2=16$ .

Hence $x^2+8x+1=0$ can be written as $x^2+8x+16-16+1=0$ or

$(x+4)^2-15=0$ which can be written as

$(x+4)^2-(sqrt15)^2=0$

Now using identity $a^2-b^2=(a+b)(a-b)$ , this can be written as

$(x+4+sqrt15)(x+4-sqrt15)=0$

Hence either $x+4+sqrt15=0$ i.e. $x=-4-sqrt15$

or $x+4-sqrt15=0$ i.e. $x=-4+sqrt15$

How do you solve x^2+8x+1=0x^2+8x+1=0 by completing the square?