What are the solutions to the equation w^2-4w+4=0 ?

The solutions to the equation w^2-4w+4=0 are w=2

Find the zeros of w^2-4w+4=0

The zeros of w^2-4w+4=0 are w=2

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w^{2} - 4 w + 4 = 0

w = 2

Solution steps

w^{2} - 4 w + 4 = 0

Solve with the quadratic formula

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

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

w_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

w = \frac{- ( - 4 )}{2 \cdot 1}

\frac{- ( - 4 )}{2 \cdot 1} = 2

w = 2

The solution to the quadratic equation is:

w = 2

0 = x^{2} - 5 x + 6

lo g_{10} (5)

lo g_{0.5} (1)

\frac{2}{- 10}

∣ 4 x - 2 ∣ + 1 = 15

What are the solutions to the equation w^2-4w+4=0 ?
The solutions to the equation w^2-4w+4=0 are w=2
Find the zeros of w^2-4w+4=0
The zeros of w^2-4w+4=0 are w=2