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

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

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Answer 1 · 2016-06-06T08:51:56

$x = - 4 - \sqrt{15}$ $x=-4-sqrt15$ or

$x = - 4 + \sqrt{15}$ $x=-4+sqrt15$

Explanation:

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

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

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

${(x + 4)}^{2} - {(\sqrt{15})}^{2} = 0$ $(x+4)^2-(sqrt15)^2=0$

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

$(x + 4 + \sqrt{15}) (x + 4 - \sqrt{15}) = 0$ $(x+4+sqrt15)(x+4-sqrt15)=0$

Hence either $x + 4 + \sqrt{15} = 0$ $x+4+sqrt15=0$ i.e. $x = - 4 - \sqrt{15}$ $x=-4-sqrt15$

or $x + 4 - \sqrt{15} = 0$ $x+4-sqrt15=0$ i.e. $x = - 4 + \sqrt{15}$ $x=-4+sqrt15$

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

1 Answer

Explanation:

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Science

Math

Humanities

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How do you solve x2+8x+1=0x^2+8x+1=0 by completing the square?

1 Answer

Explanation:

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