How do you solve $x^{2} + 6 x = 3$ $x^2 + 6x = 3$ using completing the square?

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Answer 1 · 2015-07-23T02:36:00

To complete the square you add $9$ $9$ to both sides.

Method : you halve the $x$ $x$ -coefficient ( $6$ $6$ ) to get $3$ $3$ , and the square of that ( $9$ $9$ ) is needed to complete the square.

$\to x^{2} + 6 x + 9 = 3 + 9$ $->x^2+6x+9=3+9$

$\to {(x + 3)}^{2} = 12 \to x + 3 = \pm \sqrt{12} = \pm 2 \sqrt{3}$ $->(x+3)^2=12->x+3=+-sqrt12=+-2sqrt3$

$\to x_{1, 2} = - 3 \pm 2 \sqrt{3}$ $->x_(1,2)=-3+-2sqrt3$

(remember $\pm$ $+-$ means two solutions, one with + and one with -)

How do you solve x^2 + 6x = 3x2+6x=3x^2 + 6x = 3 using completing the square?