What is 2+2x+2x^2>= 6+x+x^2 ?

The solution to 2+2x+2x^2>= 6+x+x^2 is x = (sqrt(17)-1)/2

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2+2x+2x^2>= 6+x+x^2

2 + 2 x + 2 x^{2} \geq 6 + x + x^{2}

x \leq \frac{- 17 - 1}{2} or x \geq \frac{17 - 1}{2}

Interval Notation

(- \infty, \frac{- 17 - 1}{2}] \cup [\frac{17 - 1}{2}, \infty)

Decimal Notation

x \leq - 2.56155 \dots or x \geq 1.56155 \dots

Solution steps

2 + 2 x + 2 x^{2} \geq 6 + x + x^{2}

Rewrite in standard form

x^{2} + x - 4 \geq 0

Complete the square

x^{2} + x - 4 : (x + \frac{1}{2})^{2} - \frac{17}{4}

(x + \frac{1}{2})^{2} - \frac{17}{4} \geq 0

Move

\frac{17}{4}

to the right side

(x + \frac{1}{2})^{2} \geq \frac{17}{4}

For

u^{n} \geq a

, if

n

is even then

x + \frac{1}{2} \leq - \frac{17}{4} or x + \frac{1}{2} \geq \frac{17}{4}

x + \frac{1}{2} \leq - \frac{17}{4} : x \leq \frac{- 17 - 1}{2}

x + \frac{1}{2} \geq \frac{17}{4} : x \geq \frac{17 - 1}{2}

Combine the intervals

x \leq \frac{- 17 - 1}{2} or x \geq \frac{17 - 1}{2}

7 - 8

- x + ln (1 + x) < ln (0.01)

3 (x + 4)^{2} = 21

g^{2} - 14 g + 13

\frac{2}{3} x - \frac{5}{3} \leq 3 x + \frac{1}{3} \leq 4 x - \frac{7}{3}

What is 2+2x+2x^2>= 6+x+x^2 ?
The solution to 2+2x+2x^2>= 6+x+x^2 is x<= (-sqrt(17)-1)/2 \lor x>= (sqrt(17)-1)/2