How do you solve $2 x^{2} - 6 x + 1 = 0$ $2x^2 - 6x +1 = 0$ by completing the square?

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How do you solve $2 x^{2} - 6 x + 1 = 0$ $2x^2 - 6x +1 = 0$ by completing the square?

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Answer 1 · 2017-01-21T04:52:49

$x = \sqrt{\frac{7}{4}} + \frac{3}{2}$ $x=sqrt(7/4)+3/2$

$x = - \sqrt{\frac{7}{4}} + \frac{3}{2}$ $x=-sqrt(7/4)+3/2$

Explanation:

Given -

$2 x^{2} - 6 x + 1 = 0$ $2x^2-6x+1=0$

$2 x^{2} - 6 x = - 1$ $2x^2-6x=-1$

$\frac{2 x^{2}}{2} - \frac{6 x}{2} = \frac{- 1}{2}$ $(2x^2)/2-(6x)/2=(-1)/2$

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

$x^{2} - 3 x + \frac{9}{4} = \frac{- 1}{2} + \frac{9}{4}$ $x^2-3x+9/4=(-1)/2+9/4$

${(x - \frac{3}{2})}^{2} = \frac{- 2 + 9}{4} = \frac{7}{4}$ $(x-3/2)^2=(-2+9)/4=7/4$

$(x - \frac{3}{2}) = \pm \sqrt{\frac{7}{4}}$ $(x-3/2)=+-sqrt(7/4)$

$x = \sqrt{\frac{7}{4}} + \frac{3}{2}$ $x=sqrt(7/4)+3/2$

$x = - \sqrt{\frac{7}{4}} + \frac{3}{2}$ $x=-sqrt(7/4)+3/2$

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How do you solve $2 x^{2} - 6 x + 1 = 0$ $2x^2 - 6x +1 = 0$ by completing the square?

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How do you solve $2 x^{2} - 6 x + 1 = 0$ $2x^2 - 6x +1 = 0$ by completing the square?