Solve Using the Quadratic Formula v^2-2v+1=0

$v^{2} - 2 v + 1 = 0$

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 $a = 1$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $v$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $4$ .

$v = \frac{2 \pm \sqrt{0}}{2 \cdot 1}$

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

$v = \frac{2 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$v = \frac{2 \pm 0}{2 \cdot 1}$

$2$ plus or minus $0$ is $2$ .

$v = \frac{2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$v = \frac{2}{2}$

Divide $2$ by $2$ .

$v = 1$

Step 4

The final answer is the combination of both solutions.

$v = 1$ Double roots