The solution to x^2+x+1>2 is x (sqrt(5)-1)/2

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x^2+x+1>2

Solution

x^{2} + x + 1 > 2

x < \frac{- 5 - 1}{2} or x > \frac{5 - 1}{2}

Interval Notation

(- \infty, \frac{- 5 - 1}{2}) \cup (\frac{5 - 1}{2}, \infty)

Decimal Notation

x < - 1.61803 \dots or x > 0.61803 \dots

Solution steps

x^{2} + x + 1 > 2

Rewrite in standard form

x^{2} + x - 1 > 0

Complete the square

x^{2} + x - 1 : (x + \frac{1}{2})^{2} - \frac{5}{4}

(x + \frac{1}{2})^{2} - \frac{5}{4} > 0

Move

\frac{5}{4}

to the right side

(x + \frac{1}{2})^{2} > \frac{5}{4}

For

u^{n} > a

, if

n

is even then

x + \frac{1}{2} < - \frac{5}{4} or x + \frac{1}{2} > \frac{5}{4}

x + \frac{1}{2} < - \frac{5}{4} : x < \frac{- 5 - 1}{2}

x + \frac{1}{2} > \frac{5}{4} : x > \frac{5 - 1}{2}

Combine the intervals

x < \frac{- 5 - 1}{2} or x > \frac{5 - 1}{2}

2 x^{2} + 5 x - 12

3 (6 b - 1) > 18 - 3 b

(\frac{1}{e})^{- 2}

2 x - 3 = 7

t + 16 = 7

What is x^2+x+1>2 ?
The solution to x^2+x+1>2 is x<(-sqrt(5)-1)/2 \lor x>(sqrt(5)-1)/2