How do you solve by completing the square: $x^{2} + 6 x + 1 = 0$ $x^2 + 6x + 1 = 0$ ?

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How do you solve by completing the square: $x^{2} + 6 x + 1 = 0$ $x^2 + 6x + 1 = 0$ ?

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Answer 1 · 2015-04-02T15:18:22

x= -3+2 $\sqrt{2}$ $sqrt2$ , -3-2 $\sqrt{2}$ $sqrt2$

Rewrite as $x^{2}$ $x^2$ +6x+9-9+1=0
$x^{2}$ $x^2$ +6x+9=8
${(x + 3)}^{2}$ $(x+3)^2$ = 8
x+3= 2 $\sqrt{2}$ $sqrt2$ , -2 $\sqrt{2}$ $sqrt2$
x= -3 + 2 $\sqrt{2}$ $sqrt2$ ; -3 -2 $\sqrt{2}$ $sqrt2$

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How do you solve by completing the square: $x^{2} + 6 x + 1 = 0$ $x^2 + 6x + 1 = 0$ ?

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How do you solve by completing the square: x^2 + 6x + 1 = 0x2+6x+1=0x^2 + 6x + 1 = 0?

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How do you solve by completing the square: $x^{2} + 6 x + 1 = 0$ $x^2 + 6x + 1 = 0$ ?