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

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

Find the zeros of λ^2-2λ+1=0

The zeros of λ^2-2λ+1=0 are λ=1

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λ^{2} - 2 λ + 1 = 0

λ = 1

Solution steps

λ^{2} - 2 λ + 1 = 0

Solve with the quadratic formula

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

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

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

λ = \frac{- ( - 2 )}{2 \cdot 1}

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

λ = 1

The solution to the quadratic equation is:

λ = 1

4 x^{2} + 6 x = 12

2 x (x + 1) + 9 = (x + 3)^{2}

115400 \cdot D^{2} - 251245 \cdot D - 860295 = 0

2 x^{2} - 1922 = 0

(3 x - 7)^{2} - 4 (3 x - 7) - 12 = 0

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