The solution to 3x+4 (sqrt(17)+3)/2

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3x+4<x^2+2

Solution

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

x < \frac{- 17 + 3}{2} or x > \frac{17 + 3}{2}

Interval Notation

(- \infty, \frac{- 17 + 3}{2}) \cup (\frac{17 + 3}{2}, \infty)

Decimal Notation

x < - 0.56155 \dots or x > 3.56155 \dots

Solution steps

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

Rewrite in standard form

3 x + 2 - x^{2} < 0

Complete the square

3 x + 2 - x^{2} : - (x - \frac{3}{2})^{2} + \frac{17}{4}

- (x - \frac{3}{2})^{2} + \frac{17}{4} < 0

Move

\frac{17}{4}

to the right side

- (x - \frac{3}{2})^{2} < - \frac{17}{4}

Multiply both sides by

- 1

(x - \frac{3}{2})^{2} > \frac{17}{4}

For

u^{n} > a

, if

n

is even then

x - \frac{3}{2} < - \frac{17}{4} or x - \frac{3}{2} > \frac{17}{4}

x - \frac{3}{2} < - \frac{17}{4} : x < \frac{- 17 + 3}{2}

x - \frac{3}{2} > \frac{17}{4} : x > \frac{17 + 3}{2}

Combine the intervals

x < \frac{- 17 + 3}{2} or x > \frac{17 + 3}{2}

2 x^{3} + 4 x^{2} - 24 x + 18 \geq 0

(x - 5) (x - 6) \leq 0

4 (x - 2) (x - 3) \leq 2 (x^{2} - 4)

x^{2} + 1 + (x + 1) 2 < 0

\frac{( ( x - 3 ) ( x - 5 ) )}{( x + 1 )} \leq 0

What is 3x+4<x^2+2 ?
The solution to 3x+4<x^2+2 is x<(-sqrt(17)+3)/2 \lor x>(sqrt(17)+3)/2