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

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

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Answer 1 · 2017-05-20T02:12:11

$x = 1 + \sqrt{1.5}$ $x=1+sqrt(1.5)$
$x = 1 - \sqrt{1.5}$ $x=1-sqrt(1.5)$

Explanation:

Given-

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

$\frac{2 x^{2}}{2} - \frac{4 x}{2} = \frac{1}{2}$ $(2x^2)/2-(4x)/2=1/2$
$x^{2} - 2 x = 0.5$ $x^2-2x=0.5$

$x^{2} - 2 x + 1 = 0.5 + 1 = 1.5$ $x^2-2x+1=0.5+1=1.5$
${(x - 1)}^{2} = 1.5$ $(x-1)^2=1.5$
$x - 1 = \pm \sqrt{1.5}$ $x-1=+-sqrt(1.5)$
$x = 1 \pm \sqrt{1.5}$ $x=1+-sqrt(1.5)$

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

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