What is 2x^2-3x-1<0 ?

The solution to 2x^2-3x-1<0 is (-sqrt(17)+3)/4 <x<(sqrt(17)+3)/4

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2x^2-3x-1<0

2 x^{2} - 3 x - 1 < 0

\frac{- 17 + 3}{4} < x < \frac{17 + 3}{4}

Interval Notation

(\frac{- 17 + 3}{4}, \frac{17 + 3}{4})

Decimal Notation

- 0.28077 \dots < x < 1.78077 \dots

Solution steps

2 x^{2} - 3 x - 1 < 0

Complete the square

2 x^{2} - 3 x - 1 : 2 (x - \frac{3}{4})^{2} - \frac{17}{8}

2 (x - \frac{3}{4})^{2} - \frac{17}{8} < 0

Move

\frac{17}{8}

to the right side

2 (x - \frac{3}{4})^{2} < \frac{17}{8}

Divide both sides by

2

(x - \frac{3}{4})^{2} < \frac{17}{16}

For

u^{n} < a

, if

n

is even then

- \frac{17}{16} < x - \frac{3}{4} < \frac{17}{16}

a < u < b

then

a < u and u < b

- \frac{17}{16} < x - \frac{3}{4} and x - \frac{3}{4} < \frac{17}{16}

- \frac{17}{16} < x - \frac{3}{4} : x > \frac{- 17 + 3}{4}

x - \frac{3}{4} < \frac{17}{16} : x < \frac{17 + 3}{4}

Combine the intervals

x > \frac{- 17 + 3}{4} and x < \frac{17 + 3}{4}

Merge Overlapping Intervals

\frac{- 17 + 3}{4} < x < \frac{17 + 3}{4}

9 y - 8 = 18 - 4 y

(2 m^{2} - 8 m + 2) - (- 3 m^{2} + 5 m - 7)

231

2 x + \frac{3 x - 1}{2} - \frac{5 x - 2}{3} = 2

5 x^{2} - 45 y^{2}

What is 2x^2-3x-1<0 ?
The solution to 2x^2-3x-1<0 is (-sqrt(17)+3)/4 <x<(sqrt(17)+3)/4