What is the answer to solvefor k,4x^2-2(k+1)x+(k+4)=0 ?

The answer to solvefor k,4x^2-2(k+1)x+(k+4)=0 is k=(-4x^2-4+2x)/(-2x+1);\quad x\ne 1/2

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solvefor k,4x^2-2(k+1)x+(k+4)=0

solve for

k, 4 x^{2} - 2 (k + 1) x + (k + 4) = 0

k = \frac{- 4 x ^{2} - 4 + 2 x}{- 2 x + 1}; x \neq = \frac{1}{2}

Solution steps

4 x^{2} - 2 (k + 1) x + (k + 4) = 0

4 x^{2} - 2 x (k + 1) + k + 4 = 0

Move

4 x^{2}

to the right side

- 2 x (k + 1) + k + 4 = - 4 x^{2}

Move

4

to the right side

- 2 x (k + 1) + k = - 4 x^{2} - 4

Expand

- 2 x (k + 1) : - 2 k x - 2 x

- 2 k x - 2 x + k = - 4 x^{2} - 4

Move

2 x

to the right side

- 2 k x + k = - 4 x^{2} - 4 + 2 x

Factor

- 2 k x + k : k (- 2 x + 1)

k (- 2 x + 1) = - 4 x^{2} - 4 + 2 x

Divide both sides by

- 2 x + 1; x \neq = \frac{1}{2}

k = \frac{- 4 x ^{2} - 4 + 2 x}{- 2 x + 1}; x \neq = \frac{1}{2}

\frac{n}{6} = \frac{6}{9}

x + 2 x + 3 x = 180

6 x + 3 = 3 x - 12

\frac{5 x}{3} - \frac{x - 2}{4} = \frac{9}{3} - \frac{1}{2} (x - \frac{2 x - 1}{3})

8 (1 + 2 x) + 7 x = - 153

What is the answer to solvefor k,4x^2-2(k+1)x+(k+4)=0 ?
The answer to solvefor k,4x^2-2(k+1)x+(k+4)=0 is k=(-4x^2-4+2x)/(-2x+1);\quad x\ne 1/2