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

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x^2+x>1

Solution

x^{2} + x > 1

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

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}

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

y^{3} + 1

2 n - 4 > - 12

29.3 + x \geq 10

y + 1 = 0

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