Solve Using the Quadratic Formula 4x^2-4x=-1

$4 x^{2} - 4 x = - 1$

Step 1

Add $1$ to both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $16$ .

$x = \frac{4 \pm \sqrt{0}}{2 \cdot 4}$

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

$x = \frac{4 \pm \sqrt{0^{2}}}{2 \cdot 4}$

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

$x = \frac{4 \pm 0}{2 \cdot 4}$

$4$ plus or minus $0$ is $4$ .

$x = \frac{4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{4}{8}$

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$x = \frac{4 \cdot 1}{4 \cdot 2}$

Cancel the common factor.

$x = \frac{4 \cdot 1}{4 \cdot 2}$

Rewrite the expression.

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

Step 5

The final answer is the combination of both solutions.

$x = \frac{1}{2}$ Double roots