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 \leq 0

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

Interval Notation

[\frac{- 17 + 3}{4}, \frac{17 + 3}{4}]

Decimal Notation

- 0.28077 \dots \leq x \leq 1.78077 \dots

Solution steps

2 x^{2} - 3 x - 1 \leq 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} \leq 0

Move

\frac{17}{8}

to the right side

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

Divide both sides by

2

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

For

u^{n} \leq a

, if

n

is even then

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

a \leq u \leq b

then

a \leq u and u \leq b

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

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

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

Combine the intervals

x \geq \frac{- 17 + 3}{4} and x \leq \frac{17 + 3}{4}

Merge Overlapping Intervals

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

3 x^{2} - 3 x = 11

- x < 3

\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{5 x ^{2}} - \frac{1}{5 y ^{2}}}

\frac{73}{18} = \frac{2}{3} n

\frac{x ^{3}}{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