How do you solve $2 x^{2} + 4 x - 1 = 0$ $2x^2+4x-1=0$ by completing the square?

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Answer 1 · 2017-08-31T11:51:22

$x = - 1 \pm \frac{1}{2} \sqrt{6}$ $x=-1±1/2sqrt6$

$2 x^{2} + 4 x - 1 = 0$ $2x^2+4x-1=0$

$x^{2} + 2 x - \frac{1}{2} = 0$ $x^2+2x-1/2=0$

${(x + 1)}^{2} - 1 - \frac{1}{2} = 0$ $(x+1)^2-1-1/2=0$

${(x + 1)}^{2} = \frac{3}{2}$ $(x+1)^2=3/2$

$x + 1 = \pm \frac{1}{2} \sqrt{6}$ $x+1=±1/2sqrt6$ (Note that $\sqrt{\frac{3}{2}} \equiv \frac{1}{2} \sqrt{6}$ $sqrt(3/2)-=1/2sqrt6$ .)

$x = - 1 \pm \frac{1}{2} \sqrt{6}$ $x=-1±1/2sqrt6$

How do you solve 2x^2+4x-1=02x2+4x−1=02x^2+4x-1=0 by completing the square?