How do you solve using the completing the square method $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x +1 = 0$ ?

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How do you solve using the completing the square method $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x +1 = 0$ ?

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Answer 1 · 2018-04-09T01:16:28

Please following my working.

Explanation:

$2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x + 1 = 0$
$x^{2} - \frac{3 x}{2} + \frac{1}{2} = 0$ $x^2 - (3x)/2 + 1/2 = 0$
$x^{2} - \frac{3 x}{2} + {(\frac{3}{4})}^{2} = {(\frac{3}{4})}^{2} - \frac{1}{2}$ $x^2 - (3x)/2 + (3/4)^2 = (3/4)^2 - 1/2$
${(x - \frac{3}{4})}^{2} = \frac{9}{16} - \frac{8}{16}$ $(x-3/4)^2 = 9/16 - 8/16$
${(x - \frac{3}{4})}^{2} = \frac{1}{16}$ $(x- 3/4)^2 = 1/16$
$x - \frac{3}{4} = \pm \frac{1}{4}$ $x-3/4 = +-1/4$
$x = \frac{3}{4} \pm \frac{1}{4}$ $x = 3/4+-1/4$

Therefore:
$x = 1, \frac{1}{2}$ $x = 1, 1/2$

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How do you solve using the completing the square method $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x +1 = 0$ ?

1 Answer

Explanation:

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