What is solvefor k,(k-2)x^2+8x+k+4<0 ?

The solution to solvefor k,(k-2)x^2+8x+k+4<0 is k<(-8x-4+2x^2)/(x^2+1);\quad True for all x

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solvefor k,(k-2)x^2+8x+k+4<0

solve for

k, (k - 2) x^{2} + 8 x + k + 4 < 0

k < \frac{- 8 x - 4 + 2 x ^{2}}{x ^{2} + 1}; True for all x

Solution steps

(k - 2) x^{2} + 8 x + k + 4 < 0

Move

8 x

to the right side

(k - 2) x^{2} + k + 4 < - 8 x

Move

4

to the right side

(k - 2) x^{2} + k < - 8 x - 4

Expand

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

k x^{2} - 2 x^{2} + k < - 8 x - 4

Move

2 x^{2}

to the right side

k x^{2} + k < - 8 x - 4 + 2 x^{2}

Factor

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

k (x^{2} + 1) < - 8 x - 4 + 2 x^{2}

Divide both sides by

x^{2} + 1;

True for all

x

k < \frac{- 8 x - 4 + 2 x ^{2}}{x ^{2} + 1}; True for all x

6 x - 5 \leq 13

\frac{( 5 x - 5 )}{2} > \frac{( 4 x + 20 )}{3}

\frac{( b ) ( 2 ( x - 1 ) )}{5} \leq \frac{( 3 ( 2 + x ) )}{7}

23 x - 27 - 13 x < 129 - 6

3 (x - 5) + 4 \leq 4 x - 18

What is solvefor k,(k-2)x^2+8x+k+4<0 ?
The solution to solvefor k,(k-2)x^2+8x+k+4<0 is k<(-8x-4+2x^2)/(x^2+1);\quad True for all x