The solution to x^2+5x-2<0 is (-sqrt(33)-5)/2 <x<(sqrt(33)-5)/2

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x^2+5x-2<0

Solution

x^{2} + 5 x - 2 < 0

\frac{- 33 - 5}{2} < x < \frac{33 - 5}{2}

Interval Notation

(\frac{- 33 - 5}{2}, \frac{33 - 5}{2})

Decimal Notation

- 5.37228 \dots < x < 0.37228 \dots

Solution steps

x^{2} + 5 x - 2 < 0

Complete the square

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

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

Move

\frac{33}{4}

to the right side

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

For

u^{n} < a

, if

n

is even then

- \frac{33}{4} < x + \frac{5}{2} < \frac{33}{4}

a < u < b

then

a < u and u < b

- \frac{33}{4} < x + \frac{5}{2} and x + \frac{5}{2} < \frac{33}{4}

- \frac{33}{4} < x + \frac{5}{2} : x > \frac{- 33 - 5}{2}

x + \frac{5}{2} < \frac{33}{4} : x < \frac{33 - 5}{2}

Combine the intervals

x > \frac{- 33 - 5}{2} and x < \frac{33 - 5}{2}

Merge Overlapping Intervals

\frac{- 33 - 5}{2} < x < \frac{33 - 5}{2}

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

\frac{- \frac{2}{y} - \frac{5}{x}}{- 6 x}

2 (x - 3) - 5 x < x - 2

(x^{2}) (x^{2})

6 (\frac{3}{2})^{2}

What is x^2+5x-2<0 ?
The solution to x^2+5x-2<0 is (-sqrt(33)-5)/2 <x<(sqrt(33)-5)/2