What is 2x^2-2x-1<0 ?

The solution to 2x^2-2x-1<0 is (-sqrt(3)+1)/2 <x<(sqrt(3)+1)/2

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2x^2-2x-1<0

2 x^{2} - 2 x - 1 < 0

\frac{- 3 + 1}{2} < x < \frac{3 + 1}{2}

Interval Notation

(\frac{- 3 + 1}{2}, \frac{3 + 1}{2})

Decimal Notation

- 0.36602 \dots < x < 1.36602 \dots

Solution steps

2 x^{2} - 2 x - 1 < 0

Complete the square

2 x^{2} - 2 x - 1 : 2 (x - \frac{1}{2})^{2} - \frac{3}{2}

2 (x - \frac{1}{2})^{2} - \frac{3}{2} < 0

Move

\frac{3}{2}

to the right side

2 (x - \frac{1}{2})^{2} < \frac{3}{2}

Divide both sides by

2

(x - \frac{1}{2})^{2} < \frac{3}{4}

For

u^{n} < a

, if

n

is even then

- \frac{3}{4} < x - \frac{1}{2} < \frac{3}{4}

a < u < b

then

a < u and u < b

- \frac{3}{4} < x - \frac{1}{2} and x - \frac{1}{2} < \frac{3}{4}

- \frac{3}{4} < x - \frac{1}{2} : x > \frac{- 3 + 1}{2}

x - \frac{1}{2} < \frac{3}{4} : x < \frac{3 + 1}{2}

Combine the intervals

x > \frac{- 3 + 1}{2} and x < \frac{3 + 1}{2}

Merge Overlapping Intervals

\frac{- 3 + 1}{2} < x < \frac{3 + 1}{2}

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

\frac{1}{2 ( 3 - 8 x )} > 20 (\frac{1 - 1}{5 x})

\frac{1}{( x - 2 )} \leq 0

\frac{1}{( x ^{2} + x )} \leq \frac{1}{( 2 x ^{2} + 2 x + 3 )}

\frac{2}{x - 1} \leq \frac{3}{2 x - 3}

What is 2x^2-2x-1<0 ?
The solution to 2x^2-2x-1<0 is (-sqrt(3)+1)/2 <x<(sqrt(3)+1)/2