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

The solutions to the equation solvefor x,(x+1)(x+2)-(x+2)(k-2)=0 are x=k-3,x=-2

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

The zeros of solvefor x,(x+1)(x+2)-(x+2)(k-2)=0 are x=k-3,x=-2

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x, (x + 1) (x + 2) - (x + 2) (k - 2) = 0

x = k - 3, x = - 2

Solution steps

(x + 1) (x + 2) - (x + 2) (k - 2) = 0

Expand

(x + 1) (x + 2) - (x + 2) (k - 2) : x^{2} + 5 x - k x + 6 - 2 k

x^{2} + 5 x - k x + 6 - 2 k = 0

Write in the standard form

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

x^{2} + (5 - k) x + 6 - 2 k = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 5 - k ) \pm ( 5 - k ) ^{2} - 4 \cdot 1 \cdot ( 6 - 2 k )}{2 \cdot 1}

Simplify

(5 - k)^{2} - 4 \cdot 1 \cdot (6 - 2 k) : k - 1

x_{1, 2} = \frac{- ( 5 - k ) \pm ( k - 1 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 5 - k ) + k - 1}{2 \cdot 1}, x_{2} = \frac{- ( 5 - k ) - ( k - 1 )}{2 \cdot 1}

x = \frac{- ( 5 - k ) + k - 1}{2 \cdot 1} : k - 3

x = \frac{- ( 5 - k ) - ( k - 1 )}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = k - 3, x = - 2

r^{2} + 490 = 0

256 c^{2} = 169

- 8 x^{2} - 16 x = 0

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

a, a^{2} = b^{2} \frac{1}{2}

What are the solutions to the equation solvefor x,(x+1)(x+2)-(x+2)(k-2)=0 ?
The solutions to the equation solvefor x,(x+1)(x+2)-(x+2)(k-2)=0 are x=k-3,x=-2
Find the zeros of solvefor x,(x+1)(x+2)-(x+2)(k-2)=0
The zeros of solvefor x,(x+1)(x+2)-(x+2)(k-2)=0 are x=k-3,x=-2