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Popular >

2x^2+9x-1>0

Solution

2 x^{2} + 9 x - 1 > 0

Solution

x < \frac{- 89 - 9}{4} or x > \frac{89 - 9}{4}

+2

Interval Notation

(- \infty, \frac{- 89 - 9}{4}) \cup (\frac{89 - 9}{4}, \infty)

Decimal Notation

x < - 4.60849 \dots or x > 0.10849 \dots

Solution steps

2 x^{2} + 9 x - 1 > 0

Complete the square

2 x^{2} + 9 x - 1 : 2 (x + \frac{9}{4})^{2} - \frac{89}{8}

2 (x + \frac{9}{4})^{2} - \frac{89}{8} > 0

Move

\frac{89}{8}

to the right side

2 (x + \frac{9}{4})^{2} > \frac{89}{8}

Divide both sides by

2

(x + \frac{9}{4})^{2} > \frac{89}{16}

For

u^{n} > a

, if

n

is even then

x + \frac{9}{4} < - \frac{89}{16} or x + \frac{9}{4} > \frac{89}{16}

x + \frac{9}{4} < - \frac{89}{16} : x < \frac{- 89 - 9}{4}

x + \frac{9}{4} > \frac{89}{16} : x > \frac{89 - 9}{4}

Combine the intervals

x < \frac{- 89 - 9}{4} or x > \frac{89 - 9}{4}

Popular Examples

simplify (sqrt(x+4))^2 simplify

(x + 4)^{2}

- 3 a + 1 = 1

factor u*(u-v)+v*(v-u)factor

u \cdot (u - v) + v \cdot (v - u)

0 = 3 x^{2} - 4 x - 4

- 46 x - 23 = 46 x + 23

Frequently Asked Questions (FAQ)

What is 2x^2+9x-1>0 ?
The solution to 2x^2+9x-1>0 is x<(-sqrt(89)-9)/4 \lor x>(sqrt(89)-9)/4