How do you solve $4 x^{2} - 2 x - 5 = 0$ $4x^2-2x-5=0$ by completing the square?

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How do you solve $4 x^{2} - 2 x - 5 = 0$ $4x^2-2x-5=0$ by completing the square?

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Answer 1 · 2015-04-30T00:11:57

$4 x^{2} - 2 x - 5 = 4 x^{2} - 2 x + \frac{1}{4} - 5 - \frac{1}{4} =$ $4x^2 - 2x - 5 = 4x^2 - 2x + 1/4 - 5 - 1/4 =$
${(2 x - \frac{1}{2})}^{2} = 5 + \frac{1}{4} = \frac{21}{4}$ $(2x - 1/2)^2 = 5 + 1/4 = 21/4$

a. #(2x - 1/2) = (sqr21)/2 -> 2x = 1/2 + (sqr21)/2 ->

$x = \frac{1}{4} + \frac{s q r 21}{4}$ $x = 1/4 + (sqr21)/4$

b. $(2 x - \frac{1}{2}) = - \frac{s q r 21}{2} \to 2 x = \frac{1}{2} - \frac{s q r 21}{2}$ $(2x - 1/2) = -(sqr 21)/2 -> 2x = 1/2 - (sqr21)/2$ ->

$x = \frac{1}{4} - \frac{s q r 21}{4}$ $x = 1/4 - (sqr21)/4$

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How do you solve $4 x^{2} - 2 x - 5 = 0$ $4x^2-2x-5=0$ by completing the square?

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