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

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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 · 2016-11-19T20:06:24

$x = 2.22 or x = 0225$ $x = 2.22 " or " x = 0225$

Explanation:

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

Divide both sides by 2 to get just $x^{2}$ $x^2$

$x^{2} - 2 x - \frac{1}{2} = 0 \leftarrow$ $x^2 -2x-1/2 =0" "larr$ move $- \frac{1}{2}$ $-1/2$ to the other side
$x^{2} - 2 x = \frac{1}{2} \leftarrow$ $x^2 -2x " "= 1/2" "larr$ add ${(- \frac{2}{2})}^{2}$ $color(red)((-2/2)^2)$ to both sides
$x^{2} - 2 x + 1 = \frac{1}{2} + 1$ $x^2 -2x " "color(red)(+1)= 1/2 color(red)(+1)$

${(x - 1)}^{2} = \frac{3}{2} \leftarrow$ $(x-1)^2" "=3/2" "larr$ write as square of a binomial

$x - 1 = \pm \sqrt{\frac{3}{2}} \leftarrow$ $x-1" "=+-sqrt(3/2)" "larr$ find square root of each side

$x = + \sqrt{\frac{3}{2}} + 1 or x = - \sqrt{\frac{3}{2}} + 1 \leftarrow$ $x = +sqrt(3/2)+1" "or x = -sqrt(3/2)+1" "larr$ solve for x

$x = 2.22 or x = 0225$ $x = 2.22 " or " x = 0225$

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

1 Answer

Explanation:

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