Solve for x 16x^2-8x+1=0

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

Step 1

Factor using the perfect square rule.

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

${(4 x)}^{2} - 8 x + 1 = 0$

Rewrite $1$ as $1^{2}$ .

${(4 x)}^{2} - 8 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot (4 x) \cdot 1$

Rewrite the polynomial.

${(4 x)}^{2} - 2 \cdot (4 x) \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 4 x$ and $b = 1$ .

${(4 x - 1)}^{2} = 0$

Step 2

Set the $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Step 3

Solve for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{4}$

Decimal Form:

$x = 0.25$