How do you solve $16 x^{2} - 12 x - 1 = 0$ $16x^2-12x-1=0$ by completing the square?

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Answer 1 · 2015-05-23T21:06:10

${(4 x - \frac{3}{2})}^{2} = 16 x^{2} - 12 x + \frac{9}{4}$ $(4x-3/2)^2 = 16x^2-12x+9/4$

So

$0 = 16 x^{2} - 12 x - 1 = 16 x^{2} - 12 x + \frac{9}{4} - \frac{9}{4} - 1$ $0 = 16x^2-12x-1 = 16x^2-12x+9/4-9/4-1$

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

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

Add $\frac{13}{4}$ $13/4$ to both ends to get:

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

So $4 x - \frac{3}{2} = \pm \frac{\sqrt{13}}{2}$ $4x-3/2 = +-sqrt(13)/2$

Add $\frac{3}{2}$ $3/2$ to both sides to get:

$4 x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$ $4x = 3/2+-sqrt(13)/2$

Divide both sides by $4$ $4$ to get

$x = \frac{3}{8} \pm \frac{\sqrt{13}}{8}$ $x=3/8+-sqrt(13)/8$

How do you solve 16x2−12x−1=016x^2-12x-1=0 by completing the square?