What are the solutions to the equation v^2-2v+1=0 ?

The solutions to the equation v^2-2v+1=0 are v=1

Find the zeros of v^2-2v+1=0

The zeros of v^2-2v+1=0 are v=1

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

v = 1

Solution steps

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

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

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

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

v = \frac{- ( - 2 )}{2 \cdot 1}

\frac{- ( - 2 )}{2 \cdot 1} = 1

v = 1

The solution to the quadratic equation is:

v = 1

3 (x - 1) - \frac{9 x - 1}{2} < 26

\frac{7}{2} - 3

3 x^{2} + 4 x - 20 = 0

10000 x - x^{2} = 24000000

5 < ∣ 2 x - 5 ∣ \leq 9

What are the solutions to the equation v^2-2v+1=0 ?
The solutions to the equation v^2-2v+1=0 are v=1
Find the zeros of v^2-2v+1=0
The zeros of v^2-2v+1=0 are v=1