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Popular >

2x^2+5x-1>0

Solution

2 x^{2} + 5 x - 1 > 0

Solution

x < \frac{- 33 - 5}{4} or x > \frac{33 - 5}{4}

+2

Solution steps

2 x^{2} + 5 x - 1 > 0

Complete the square

2 x^{2} + 5 x - 1 : 2 (x + \frac{5}{4})^{2} - \frac{33}{8}

2 (x + \frac{5}{4})^{2} - \frac{33}{8} > 0

Move

\frac{33}{8}

to the right side

2 (x + \frac{5}{4})^{2} > \frac{33}{8}

Divide both sides by

2

(x + \frac{5}{4})^{2} > \frac{33}{16}

For

u^{n} > a

, if

n

is even then

x + \frac{5}{4} < - \frac{33}{16} or x + \frac{5}{4} > \frac{33}{16}

x + \frac{5}{4} < - \frac{33}{16} : x < \frac{- 33 - 5}{4}

x + \frac{5}{4} > \frac{33}{16} : x > \frac{33 - 5}{4}

Combine the intervals

x < \frac{- 33 - 5}{4} or x > \frac{33 - 5}{4}

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Frequently Asked Questions (FAQ)

What is 2x^2+5x-1>0 ?
The solution to 2x^2+5x-1>0 is x<(-sqrt(33)-5)/4 \lor x>(sqrt(33)-5)/4