Solve Using the Quadratic Formula 64x^2-16x+1=0

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

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 64$ , $b = - 16$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (64 \cdot 1)}}{2 \cdot 64}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 64 \cdot 1}}{2 \cdot 64}$

Multiply $- 4 \cdot 64 \cdot 1$ .

Multiply $- 4$ by $64$ .

$x = \frac{16 \pm \sqrt{256 - 256 \cdot 1}}{2 \cdot 64}$

Multiply $- 256$ by $1$ .

$x = \frac{16 \pm \sqrt{256 - 256}}{2 \cdot 64}$

Subtract $256$ from $256$ .

$x = \frac{16 \pm \sqrt{0}}{2 \cdot 64}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{16 \pm \sqrt{0^{2}}}{2 \cdot 64}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{16 \pm 0}{2 \cdot 64}$

$16$ plus or minus $0$ is $16$ .

$x = \frac{16}{2 \cdot 64}$

Multiply $2$ by $64$ .

$x = \frac{16}{128}$

Cancel the common factor of $16$ and $128$ .

Factor $16$ out of $16$ .

$x = \frac{16 (1)}{128}$

Cancel the common factors.

Factor $16$ out of $128$ .

$x = \frac{16 \cdot 1}{16 \cdot 8}$

Cancel the common factor.

$x = \frac{16 \cdot 1}{16 \cdot 8}$

Rewrite the expression.

$x = \frac{1}{8}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{8}$ Double roots