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

The solutions to the equation kx^2+(k-1)x-1=0 are x= 1/k ,x=-1;\quad k\ne 0

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

The zeros of kx^2+(k-1)x-1=0 are x= 1/k ,x=-1;\quad k\ne 0

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

x = \frac{1}{k}, x = - 1; k \neq = 0

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- ( k - 1 ) \pm ( k + 1 )}{2 k}; k \neq = 0

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1}{k}, x = - 1; k \neq = 0

- 5 c^{2} + 4 = - 81

- x^{2} - 3 x + 8 = 0

u (2 u - 1) a + (u - 1) (2 u - 1) b + 4 u (1 - u) c = 0

0.2 = 6 \cdot (9.5 - 0.2 \cdot x) \cdot (1 + 0.05 \cdot (9.5 - 0.2 \cdot x))

0 = x^{2} - 10 x + 16

What are the solutions to the equation kx^2+(k-1)x-1=0 ?
The solutions to the equation kx^2+(k-1)x-1=0 are x= 1/k ,x=-1;\quad k\ne 0
Find the zeros of kx^2+(k-1)x-1=0
The zeros of kx^2+(k-1)x-1=0 are x= 1/k ,x=-1;\quad k\ne 0