How do you solve by completing the square: $x^{2} + 6 x + 4 = 0$ $x^2+6x+4=0$ ?

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How do you solve by completing the square: $x^{2} + 6 x + 4 = 0$ $x^2+6x+4=0$ ?

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Answer 1 · 2015-04-03T09:44:59

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
${(x + a)}^{2} = b$ $(x+a)^2=b$
So, if $b \geq 0$ $b\ge 0$ , you can take the square root at both sides to get
$x + a = \pm \sqrt{b}$ $x+a=\pm\sqrt{b}$
and conclude $x = \pm \sqrt{b} - a$ $x=\pm\sqrt{b}-a$ .

Now, we have ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ . Since you equation starts with $x^{2} + 6 x$ $x^2+6x$ , this means that $2 a x = 6 x$ $2ax=6x$ , and so $a = 3$ $a=3$ .
Adding $5$ $5$ at both sides, we have
$x^{2} + 6 x + 9 = 5$ $x^2+6x+9=5$
Which is the form we wanted, because now we have
${(x + 3)}^{2} = 5$ $(x+3)^2=5$
Which leads us to
$x + 3 = \pm \sqrt{5}$ $x+3=\pm\sqrt{5}$ and finally $x = \pm \sqrt{5} - 3$ $x=\pm\sqrt{5}-3$

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How do you solve by completing the square: $x^{2} + 6 x + 4 = 0$ $x^2+6x+4=0$ ?

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