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

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

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

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

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

x = k + 1, x = k - 2

Solution steps

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

Expand

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(1 - 2 k)^{2} - 4 \cdot 1 \cdot (k^{2} - k - 2) = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = k + 1, x = k - 2

x^{2} - 13 x - 420 = 0

x, p (x) = y (x^{2} + 2) - x (y^{2} + 2)

- X^{2} - 2 X + 1 = 0

x, f^{- 1} (x) = 2 - (x - 3)^{2}

1 x^{2} - 7 x + 10 = 0

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