What are the solutions to the equation 16x^2-4x+1/4 =0 ?

The solutions to the equation 16x^2-4x+1/4 =0 are x= 1/8

Find the zeros of 16x^2-4x+1/4 =0

The zeros of 16x^2-4x+1/4 =0 are x= 1/8

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16 x^{2} - 4 x + \frac{1}{4} = 0

x = \frac{1}{8}

Decimal Notation

x = 0.125

Solution steps

16 x^{2} - 4 x + \frac{1}{4} = 0

Multiply both sides by

4

64 x^{2} - 16 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 64 \cdot 1}{2 \cdot 64}

(- 16)^{2} - 4 \cdot 64 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 16 ) \pm 0}{2 \cdot 64}

x = \frac{- ( - 16 )}{2 \cdot 64}

\frac{- ( - 16 )}{2 \cdot 64} = \frac{1}{8}

x = \frac{1}{8}

The solution to the quadratic equation is:

x = \frac{1}{8}

1 - 2 x > 0

\frac{( \frac{7}{25} )}{( - \frac{24}{25} )}

\frac{20 + 216}{2}

\frac{1}{3} x^{2} + \frac{1}{15} x - 1 = 0

\frac{2}{x + 1} \leq \frac{3}{x - 1}

What are the solutions to the equation 16x^2-4x+1/4 =0 ?
The solutions to the equation 16x^2-4x+1/4 =0 are x= 1/8
Find the zeros of 16x^2-4x+1/4 =0
The zeros of 16x^2-4x+1/4 =0 are x= 1/8