What are the solutions to the equation 2k(k-4)=1 ?

The solutions to the equation 2k(k-4)=1 are k=(4+3sqrt(2))/2 ,k=(4-3sqrt(2))/2

Find the zeros of 2k(k-4)=1

The zeros of 2k(k-4)=1 are k=(4+3sqrt(2))/2 ,k=(4-3sqrt(2))/2

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2 k (k - 4) = 1

k = \frac{4 + 3 2}{2}, k = \frac{4 - 3 2}{2}

Decimal Notation

k = 4.12132 \dots, k = - 0.12132 \dots

Solution steps

2 k (k - 4) = 1

Expand

2 k (k - 4) : 2 k^{2} - 8 k

2 k^{2} - 8 k = 1

Move

1

to the left side

2 k^{2} - 8 k - 1 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 1) = 62

k_{1, 2} = \frac{- ( - 8 ) \pm 6 2}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- ( - 8 ) + 6 2}{2 \cdot 2}, k_{2} = \frac{- ( - 8 ) - 6 2}{2 \cdot 2}

k = \frac{- ( - 8 ) + 6 2}{2 \cdot 2} : \frac{4 + 3 2}{2}

k = \frac{- ( - 8 ) - 6 2}{2 \cdot 2} : \frac{4 - 3 2}{2}

The solutions to the quadratic equation are:

k = \frac{4 + 3 2}{2}, k = \frac{4 - 3 2}{2}

x, z^{2} = x^{2} + y^{2}

x^{2} - 6 x - 20 = 20

4 x^{2} - 3 x - 12 = 3 x^{2} - 6 x - 14

(x - 7)^{2} = - 9

3 x^{2} - 8 x - 16

What are the solutions to the equation 2k(k-4)=1 ?
The solutions to the equation 2k(k-4)=1 are k=(4+3sqrt(2))/2 ,k=(4-3sqrt(2))/2
Find the zeros of 2k(k-4)=1
The zeros of 2k(k-4)=1 are k=(4+3sqrt(2))/2 ,k=(4-3sqrt(2))/2