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

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x^2-1>x

Solution

x^{2} - 1 > x

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

Solution steps

x^{2} - 1 > x

Rewrite in standard form

x^{2} - 1 - x > 0

Complete the square

x^{2} - 1 - x : (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} + 8 x^{2} + 4 x > 32

(7^{2})^{2}

1 - 5

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