How do you solve $6 x = 4 x^{2} - 1$ $6x = 4x^2 -1$ by completing the square?

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Answer 1 · 2017-08-24T12:13:26

$x = \frac{1}{4} (3 \pm \sqrt{13})$ $x = 1/4(3 +-sqrt13)$

$6 x = 4 x^{2} - 1 or 4 x^{2} - 6 x = 1 or 4 (x^{2} - \frac{3}{2} x) = 1$ $6x = 4x^2 -1 or 4x^2 -6x =1 or 4(x^2-3/2x) =1$ or

$4 (x^{2} - \frac{3}{2} x + \frac{9}{16}) = 1 + \frac{9}{4}$ $4(x^2-3/2x + 9/16) =1 +9/4$ [Add 9/4 on both side] or

$4 {(x - \frac{3}{4})}^{2} = \frac{13}{4} or {(x - \frac{3}{4})}^{2} = \frac{13}{16}$ $4(x-3/4)^2 = 13/4 or (x-3/4)^2 = 13/16$ or

$(x - \frac{3}{4}) = \pm \sqrt{\frac{13}{16}} or x = \frac{3}{4} \pm \frac{\sqrt{13}}{4}$ $(x-3/4) = +- sqrt(13/16 ) or x = 3/4 +- sqrt13/4$

$or x = \frac{1}{4} (3 \pm \sqrt{13})$ $or x = 1/4(3 +-sqrt13)$ [Ans]

How do you solve 6x=4x2−16x = 4x^2 -1 by completing the square?