What are the solutions to the equation 2k^2+4k+1=0 ?

The solutions to the equation 2k^2+4k+1=0 are k=(-2+sqrt(2))/2 ,k=-(sqrt(2)+1)/(sqrt(2))

Find the zeros of 2k^2+4k+1=0

The zeros of 2k^2+4k+1=0 are k=(-2+sqrt(2))/2 ,k=-(sqrt(2)+1)/(sqrt(2))

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2 k^{2} + 4 k + 1 = 0

k = \frac{- 2 + 2}{2}, k = - \frac{2 + 1}{2}

Decimal Notation

k = - 0.29289 \dots, k = - 1.70710 \dots

Solution steps

2 k^{2} + 4 k + 1 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

4^{2} - 4 \cdot 2 \cdot 1 = 22

k_{1, 2} = \frac{- 4 \pm 2 2}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- 4 + 2 2}{2 \cdot 2}, k_{2} = \frac{- 4 - 2 2}{2 \cdot 2}

k = \frac{- 4 + 2 2}{2 \cdot 2} : \frac{- 2 + 2}{2}

k = \frac{- 4 - 2 2}{2 \cdot 2} : - \frac{2 + 1}{2}

The solutions to the quadratic equation are:

k = \frac{- 2 + 2}{2}, k = - \frac{2 + 1}{2}

lo g_{b} (b)

∣ - 7 ∣ + 5

\frac{e ^{x} - 1}{∣ x + 1 ∣} < 0

(x - 1)^{2} - 10 = 8

(25)^{- \frac{1}{2}}

What are the solutions to the equation 2k^2+4k+1=0 ?
The solutions to the equation 2k^2+4k+1=0 are k=(-2+sqrt(2))/2 ,k=-(sqrt(2)+1)/(sqrt(2))
Find the zeros of 2k^2+4k+1=0
The zeros of 2k^2+4k+1=0 are k=(-2+sqrt(2))/2 ,k=-(sqrt(2)+1)/(sqrt(2))