What are the solutions to the equation solvefor x,x^2-kx+4=0 ?

The solutions to the equation solvefor x,x^2-kx+4=0 are x=(k+sqrt(k^2-16))/2 ,x=(k-sqrt(k^2-16))/2

Find the zeros of solvefor x,x^2-kx+4=0

The zeros of solvefor x,x^2-kx+4=0 are x=(k+sqrt(k^2-16))/2 ,x=(k-sqrt(k^2-16))/2

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x, x^{2} - k x + 4 = 0

x = \frac{k + k ^{2} - 16}{2}, x = \frac{k - k ^{2} - 16}{2}

Solution steps

x^{2} - k x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- k)^{2} - 4 \cdot 1 \cdot 4 : k^{2} - 16

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} - 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - k ) + k ^{2} - 16}{2 \cdot 1}, x_{2} = \frac{- ( - k ) - k ^{2} - 16}{2 \cdot 1}

x = \frac{- ( - k ) + k ^{2} - 16}{2 \cdot 1} : \frac{k + k ^{2} - 16}{2}

x = \frac{- ( - k ) - k ^{2} - 16}{2 \cdot 1} : \frac{k - k ^{2} - 16}{2}

The solutions to the quadratic equation are:

x = \frac{k + k ^{2} - 16}{2}, x = \frac{k - k ^{2} - 16}{2}

x^{2} - 1 x

2 x (x + 1) = 12 - 2 (x - 9)

x^{2} - x - 36 = 0

x, 2 x^{2} + k x + 3 = 0

- 3 x^{2} + 4 x - 59 = - 4 x^{2}

What are the solutions to the equation solvefor x,x^2-kx+4=0 ?
The solutions to the equation solvefor x,x^2-kx+4=0 are x=(k+sqrt(k^2-16))/2 ,x=(k-sqrt(k^2-16))/2
Find the zeros of solvefor x,x^2-kx+4=0
The zeros of solvefor x,x^2-kx+4=0 are x=(k+sqrt(k^2-16))/2 ,x=(k-sqrt(k^2-16))/2