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

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

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Answer 1 · 2018-05-24T18:18:04

$x = \frac{\sqrt{5} - 1}{4}$ $x=(sqrt(5)-1)/4$ or $x = - \frac{\sqrt{5} + 1}{4}$ $x=-(sqrt(5)+1)/4$

Explanation:

Dividing by $4$ $4$ we get
$x^{2} + \frac{1}{2} x - \frac{1}{4} = 0$ $x^2+1/2x-1/4=0$
completing the square
${(x + \frac{1}{4})}^{2} = \frac{5}{16}$ $(x+1/4)^2=5/16$
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How do you solve $4 x^{2} + 2 x - 1 = 0$ $4x^2 + 2x - 1 = 0$ by completing the square?

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