Find the Perfect Square Trinomial x^2-1/4x+c

$x^{2} - \frac{1}{4} x + c$

Step 1

Combine $x$ and $\frac{1}{4}$ .

$x^{2} - \frac{x}{4} + c$

Step 2

Move $- \frac{x}{4}$ .

$x^{2} + c - \frac{x}{4}$

Step 3

To find the $c$ value $c = {(\frac{b}{2})}^{2}$ , divide the coefficient of $x$ by $2$ and square the result.

Multiply the numerator by the reciprocal of the denominator.

${(- \frac{1}{4} \cdot \frac{1}{2})}^{2}$

Multiply $- \frac{1}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{1}{4}$ .

${(- \frac{1}{2 \cdot 4})}^{2}$

Multiply $2$ by $4$ .

${(- \frac{1}{8})}^{2}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{8}$ .

${(- 1)}^{2} {(\frac{1}{8})}^{2}$

Apply the product rule to $\frac{1}{8}$ .

${(- 1)}^{2} \frac{1^{2}}{8^{2}}$

Raise $- 1$ to the power of $2$ .

$1 \frac{1^{2}}{8^{2}}$

Multiply $\frac{1^{2}}{8^{2}}$ by $1$ .

$\frac{1^{2}}{8^{2}}$

One to any power is one.

$\frac{1}{8^{2}}$

Raise $8$ to the power of $2$ .

$\frac{1}{64}$

Step 4

Combine $x$ and $\frac{1}{4}$ .

$x^{2} - \frac{x}{4} + c + \frac{1}{64}$