How do you solve $8 x (2 x - 1) = - 1$ $8x(2x-1)=-1$ using the quadratic formula?

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Answer 1 · 2018-03-14T21:18:33

The answer is $x = \frac{1}{4}$ $x=1/4$ .

Expand the multiplication using the distributive property:

$8 x (2 x - 1) = - 1$ $8x(2x-1)=-1$

$16 x^{2} - 8 x = - 1$ $16x^2-8x=-1$

$16 x^{2} - 8 x + 1 = 0$ $16x^2-8x+1=0$

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

$x = \frac{8 \pm \sqrt{64 - 64}}{32}$ $color(white)x=(8+-sqrt(64-64))/(32)$

$x = \frac{8 \pm \sqrt{0}}{32}$ $color(white)x=(8+-sqrt(0))/(32)$

$x = \frac{8 \pm 0}{32}$ $color(white)x=(8+-0)/(32)$

$x = \frac{8}{32}$ $color(white)x=8/32$

$x = \frac{1}{4}$ $color(white)x=1/4$

That is the solution, hope this helped!

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