What are the solutions to the equation 2x^2-(k-1)x-k=0 ?

The solutions to the equation 2x^2-(k-1)x-k=0 are x=(k-1+sqrt(k^2+6k+1))/4 ,x=(k-1-sqrt(k^2+6k+1))/4

Find the zeros of 2x^2-(k-1)x-k=0

The zeros of 2x^2-(k-1)x-k=0 are x=(k-1+sqrt(k^2+6k+1))/4 ,x=(k-1-sqrt(k^2+6k+1))/4

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2 x^{2} - (k - 1) x - k = 0

x = \frac{k - 1 + k ^{2} + 6 k + 1}{4}, x = \frac{k - 1 - k ^{2} + 6 k + 1}{4}

Solution steps

2 x^{2} - (k - 1) x - k = 0

Expand

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

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

Write in the standard form

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

2 x^{2} + (- k + 1) x - k = 0

Solve with the quadratic formula

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

Simplify

(- k + 1)^{2} - 4 \cdot 2 (- k) : k^{2} + 6 k + 1

x_{1, 2} = \frac{- ( - k + 1 ) \pm k ^{2} + 6 k + 1}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - k + 1 ) + k ^{2} + 6 k + 1}{2 \cdot 2}, x_{2} = \frac{- ( - k + 1 ) - k ^{2} + 6 k + 1}{2 \cdot 2}

x = \frac{- ( - k + 1 ) + k ^{2} + 6 k + 1}{2 \cdot 2} : \frac{k - 1 + k ^{2} + 6 k + 1}{4}

x = \frac{- ( - k + 1 ) - k ^{2} + 6 k + 1}{2 \cdot 2} : \frac{k - 1 - k ^{2} + 6 k + 1}{4}

The solutions to the quadratic equation are:

x = \frac{k - 1 + k ^{2} + 6 k + 1}{4}, x = \frac{k - 1 - k ^{2} + 6 k + 1}{4}

3 y^{2} = 81

x, 4 x^{2} y + 1 = 9

x^{2} + 11 x + 15 = 0

4 \cdot (2 x + 5) + x \cdot (2 x + 5) = 0

(m + 6) (m - 3) = - 18

What are the solutions to the equation 2x^2-(k-1)x-k=0 ?
The solutions to the equation 2x^2-(k-1)x-k=0 are x=(k-1+sqrt(k^2+6k+1))/4 ,x=(k-1-sqrt(k^2+6k+1))/4
Find the zeros of 2x^2-(k-1)x-k=0
The zeros of 2x^2-(k-1)x-k=0 are x=(k-1+sqrt(k^2+6k+1))/4 ,x=(k-1-sqrt(k^2+6k+1))/4