What are the solutions to the equation 2k(1)+k^2=8 ?

The solutions to the equation 2k(1)+k^2=8 are k=2,k=-4

Find the zeros of 2k(1)+k^2=8

The zeros of 2k(1)+k^2=8 are k=2,k=-4

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2 k (1) + k^{2} = 8

k = 2, k = - 4

Solution steps

2 k (1) + k^{2} = 8

Expand

2 k (1) + k^{2} : 2 k + k^{2}

2 k + k^{2} = 8

Move

8

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

k = \frac{- 2 + 6}{2 \cdot 1} : 2

k = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

k = 2, k = - 4

x, f (x + 2) = - x^{2} + 4 x + 1

x (3 x - 7) = - 4

9 z^{2} - 3 z = 20

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

4 r^{2} + 3 r + 6 = 357

What are the solutions to the equation 2k(1)+k^2=8 ?
The solutions to the equation 2k(1)+k^2=8 are k=2,k=-4
Find the zeros of 2k(1)+k^2=8
The zeros of 2k(1)+k^2=8 are k=2,k=-4