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

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

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

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

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

x = - 1, x = \frac{2}{1 - k}; k \neq = 1

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- k - 1)^{2} - 4 (1 - k) (- 2) : k - 3

x_{1, 2} = \frac{- ( - k - 1 ) \pm ( k - 3 )}{2 ( 1 - k )}; k \neq = 1

Separate the solutions

x_{1} = \frac{- ( - k - 1 ) + k - 3}{2 ( 1 - k )}, x_{2} = \frac{- ( - k - 1 ) - ( k - 3 )}{2 ( 1 - k )}

x = \frac{- ( - k - 1 ) + k - 3}{2 ( 1 - k )} : - 1

x = \frac{- ( - k - 1 ) - ( k - 3 )}{2 ( 1 - k )} : \frac{2}{1 - k}

The solutions to the quadratic equation are:

x = - 1, x = \frac{2}{1 - k}; k \neq = 1

(x + 2) (x - 3) = 6

x, x^{2} y + x y^{2} = 6

4 x^{2} + 2 = 38

- 6 x^{2} + 12 x = 7

6 x^{2} - 17 x + 10 = 0

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