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

The solutions to the equation (k-2)(k-1)=-1 are k= 3/2+i(sqrt(3))/2 ,k= 3/2-i(sqrt(3))/2

Find the zeros of (k-2)(k-1)=-1

The zeros of (k-2)(k-1)=-1 are k= 3/2+i(sqrt(3))/2 ,k= 3/2-i(sqrt(3))/2

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(k - 2) (k - 1) = - 1

k = \frac{3}{2} + i \frac{3}{2}, k = \frac{3}{2} - i \frac{3}{2}

Solution steps

(k - 2) (k - 1) = - 1

Expand

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

k^{2} - 3 k + 2 = - 1

Move

1

to the left side

k^{2} - 3 k + 3 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 3 : 3 i

k_{1, 2} = \frac{- ( - 3 ) \pm 3 i}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 3 ) + 3 i}{2 \cdot 1}, k_{2} = \frac{- ( - 3 ) - 3 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

k = \frac{3}{2} + i \frac{3}{2}, k = \frac{3}{2} - i \frac{3}{2}

(- 33)^{2}

- \frac{11}{4} (\frac{11 x}{3} - \frac{15}{4}) \geq \frac{415}{8} + x

1225

x^{3} - 3 x = 0

t - \frac{2}{3} = \frac{5}{6}

What are the solutions to the equation (k-2)(k-1)=-1 ?
The solutions to the equation (k-2)(k-1)=-1 are k= 3/2+i(sqrt(3))/2 ,k= 3/2-i(sqrt(3))/2
Find the zeros of (k-2)(k-1)=-1
The zeros of (k-2)(k-1)=-1 are k= 3/2+i(sqrt(3))/2 ,k= 3/2-i(sqrt(3))/2