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

The solutions to the equation solvefor x,2x^2-kx+4=0 are x=(k+sqrt(k^2-32))/4 ,x=(k-sqrt(k^2-32))/4

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

The zeros of solvefor x,2x^2-kx+4=0 are x=(k+sqrt(k^2-32))/4 ,x=(k-sqrt(k^2-32))/4

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

x = \frac{k + k ^{2} - 32}{4}, x = \frac{k - k ^{2} - 32}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- k)^{2} - 4 \cdot 2 \cdot 4 : k^{2} - 32

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} - 32}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - k ) + k ^{2} - 32}{2 \cdot 2}, x_{2} = \frac{- ( - k ) - k ^{2} - 32}{2 \cdot 2}

x = \frac{- ( - k ) + k ^{2} - 32}{2 \cdot 2} : \frac{k + k ^{2} - 32}{4}

x = \frac{- ( - k ) - k ^{2} - 32}{2 \cdot 2} : \frac{k - k ^{2} - 32}{4}

The solutions to the quadratic equation are:

x = \frac{k + k ^{2} - 32}{4}, x = \frac{k - k ^{2} - 32}{4}

5 x^{2} = 315

4 x^{2} - 14 x^{2} + 2 = 0

3 x^{2} + 5 x + 13 = 4 x^{2} + 4 x + 1

x^{2} - 7 x - x = 0

x^{2} + 0.4 x + 4 = 0

What are the solutions to the equation solvefor x,2x^2-kx+4=0 ?
The solutions to the equation solvefor x,2x^2-kx+4=0 are x=(k+sqrt(k^2-32))/4 ,x=(k-sqrt(k^2-32))/4
Find the zeros of solvefor x,2x^2-kx+4=0
The zeros of solvefor x,2x^2-kx+4=0 are x=(k+sqrt(k^2-32))/4 ,x=(k-sqrt(k^2-32))/4