The solution to x^2-x-1<0 is (-sqrt(5)+1)/2 <x<(sqrt(5)+1)/2

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x^2-x-1<0

Solution

x^{2} - x - 1 < 0

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

Solution steps

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

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

a < u < b

then

a < u and u < b

- \frac{5}{4} < x - \frac{1}{2} and x - \frac{1}{2} < \frac{5}{4}

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

Merge Overlapping Intervals

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

e^{4 t}

3 x^{2} + 25 x + 42 = 0

5 x^{2} - 23 x - 10 = 0

lo g_{2} (x) \geq 1 + lo g_{2} (2 - x)

2 x^{2} + 8 x - 6 = 0

What is x^2-x-1<0 ?
The solution to x^2-x-1<0 is (-sqrt(5)+1)/2 <x<(sqrt(5)+1)/2