What is x^2+2x-8=<6x-x^2+1 ?

The solution to x^2+2x-8=<6x-x^2+1 is -sqrt(11/2)+1<x<sqrt(11/2)+1

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x^2+2x-8=<6x-x^2+1

x^{2} + 2 x - 8 =< 6 x - x^{2} + 1

- \frac{11}{2} + 1 < x < \frac{11}{2} + 1

Interval Notation

(- \frac{11}{2} + 1, \frac{11}{2} + 1)

Decimal Notation

- 1.34520 \dots < x < 3.34520 \dots

Solution steps

x^{2} + 2 x - 8 < 6 x - x^{2} + 1

Rewrite in standard form

2 x^{2} - 4 x - 9 < 0

Complete the square

2 x^{2} - 4 x - 9 : 2 (x - 1)^{2} - 11

2 (x - 1)^{2} - 11 < 0

Move

11

to the right side

2 (x - 1)^{2} < 11

Divide both sides by

2

(x - 1)^{2} < \frac{11}{2}

For

u^{n} < a

, if

n

is even then

- \frac{11}{2} < x - 1 < \frac{11}{2}

a < u < b

then

a < u and u < b

- \frac{11}{2} < x - 1 and x - 1 < \frac{11}{2}

- \frac{11}{2} < x - 1 : x > - \frac{11}{2} + 1

x - 1 < \frac{11}{2} : x < \frac{11}{2} + 1

Combine the intervals

x > - \frac{11}{2} + 1 and x < \frac{11}{2} + 1

Merge Overlapping Intervals

- \frac{11}{2} + 1 < x < \frac{11}{2} + 1

2 x^{2} + 2 x + 1 < \frac{15}{x ^{2} + x + 1}

\frac{1}{( x - 1 )} < 2

3 x^{2} + 2 x + 2 < 2 x^{2} + x + 4

\frac{5}{x - 3} < 1

\frac{2}{( x - 1 )} \leq \frac{2}{( 2 x - 3 )}

What is x^2+2x-8=<6x-x^2+1 ?
The solution to x^2+2x-8=<6x-x^2+1 is -sqrt(11/2)+1<x<sqrt(11/2)+1