Solve for q 4q^2-4q+1=0

$4 q^{2} - 4 q + 1 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $4 q^{2}$ as ${(2 q)}^{2}$ .

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

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

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

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

$4 q = 2 \cdot (2 q) \cdot 1$

Rewrite the polynomial.

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

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

${(2 q - 1)}^{2} = 0$

Step 2

Set the $2 q - 1$ equal to $0$ .

$2 q - 1 = 0$

Step 3

Solve for $q$ .

Add $1$ to both sides of the equation.

$2 q = 1$

Divide each term in $2 q = 1$ by $2$ and simplify.

Divide each term in $2 q = 1$ by $2$ .

$\frac{2 q}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 q}{2} = \frac{1}{2}$

Divide $q$ by $1$ .

$q = \frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$q = \frac{1}{2}$

Decimal Form:

$q = 0.5$