What is (2x^2-x-3)/8 >= 1 ?

The solution to (2x^2-x-3)/8 >= 1 is x = (sqrt(89)+1)/4

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(2x^2-x-3)/8 >= 1

\frac{2 x ^{2} - x - 3}{8} \geq 1

x \leq \frac{- 89 + 1}{4} or x \geq \frac{89 + 1}{4}

Interval Notation

(- \infty, \frac{- 89 + 1}{4}] \cup [\frac{89 + 1}{4}, \infty)

Decimal Notation

x \leq - 2.10849 \dots or x \geq 2.60849 \dots

Solution steps

\frac{2 x ^{2} - x - 3}{8} \geq 1

Rewrite in standard form

2 x^{2} - x - 11 \geq 0

Complete the square

2 x^{2} - x - 11 : 2 (x - \frac{1}{4})^{2} - \frac{89}{8}

2 (x - \frac{1}{4})^{2} - \frac{89}{8} \geq 0

Move

\frac{89}{8}

to the right side

2 (x - \frac{1}{4})^{2} \geq \frac{89}{8}

Divide both sides by

2

(x - \frac{1}{4})^{2} \geq \frac{89}{16}

For

u^{n} \geq a

, if

n

is even then

x - \frac{1}{4} \leq - \frac{89}{16} or x - \frac{1}{4} \geq \frac{89}{16}

x - \frac{1}{4} \leq - \frac{89}{16} : x \leq \frac{- 89 + 1}{4}

x - \frac{1}{4} \geq \frac{89}{16} : x \geq \frac{89 + 1}{4}

Combine the intervals

x \leq \frac{- 89 + 1}{4} or x \geq \frac{89 + 1}{4}

\frac{10 x}{( x ^{2} + 9 )} \leq 1

\frac{2}{x - 3} < 0

x - > 2 (\frac{2}{( x - 1 )})

\frac{( x + 5 )}{( x ^{2} - 3 x - 4 )} > 0

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

What is (2x^2-x-3)/8 >= 1 ?
The solution to (2x^2-x-3)/8 >= 1 is x<= (-sqrt(89)+1)/4 \lor x>= (sqrt(89)+1)/4