Respuesta :

16x² + 1 = 8x   First, make this a quadratic equation. Subtract 8x from both sides
16x² -8x + 1 = 0   Now, since ax² + bx + c = 0 is quadratic equation form,  a = 16, b                             = -8, and c = 1. Plug those into the quadratic formula.
x = [tex] \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} [/tex]   Plug in your numbers
x = [tex] \frac{-(-8) \pm \sqrt{(-8)^2 - 4(16)(1)} }{2(16)} [/tex]   Simplify the double negative (- (-8))
x = [tex] \frac{8 \pm \sqrt{(-8)^2 - 4(16)(1)} }{2(16)} [/tex]   Simplify (-8)²
x = [tex] \frac{8 \pm \sqrt{64 - 4(16)(1)} }{2(16)} [/tex]   Simplify 4(16)(1)
x = [tex] \frac{8 \pm \sqrt{64 - 64} }{2(16)} [/tex]   Subtract 64 from 64
x = [tex] \frac{8 \pm \sqrt{0} }{2(16)} [/tex]   Mutiply 2 and 16
x = [tex] \frac{8 \pm \sqrt{0} }{32} [/tex]   Get rid of the [tex] \sqrt{0} [/tex]
x = [tex] \frac{8}{32} [/tex]   Simplify
x = [tex] \frac{1}{4} [/tex]


minus 8x both sides
16x^2-8x+1=0
for
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
a=16
b=-8
c=1
x=[tex] \frac{-(-8)+/- \sqrt{(-8)^2-4(16)(1)} }{2(16)} [/tex]
x=[tex] \frac{8+/- \sqrt{64-64} }{32} [/tex]
x=[tex] \frac{8+/- \sqrt{0} }{32} [/tex]
x=[tex] \frac{8+/-0 }{32} [/tex]
x=[tex] \frac{8}{32} [/tex]
x=[tex] \frac{4}{16} [/tex]
x=[tex] \frac{2}{8} [/tex]
x=[tex] \frac{1}{4} [/tex]
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