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

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

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

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

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

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

Solution steps

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

Move

1

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{2} - 1200 = 0

5 x^{2} = 2 x + 7

5 x^{2} - 16 x - 3 = 0

- x^{2} - 4 x + 7

- x^{2} - 4 x + 4

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