The solution to x^2+5x-1<0 is (-sqrt(29)-5)/2 <x<(sqrt(29)-5)/2

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x^2+5x-1<0

Solution

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

\frac{- 29 - 5}{2} < x < \frac{29 - 5}{2}

Interval Notation

(\frac{- 29 - 5}{2}, \frac{29 - 5}{2})

Decimal Notation

- 5.19258 \dots < x < 0.19258 \dots

Solution steps

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

Complete the square

x^{2} + 5 x - 1 : (x + \frac{5}{2})^{2} - \frac{29}{4}

(x + \frac{5}{2})^{2} - \frac{29}{4} < 0

Move

\frac{29}{4}

to the right side

(x + \frac{5}{2})^{2} < \frac{29}{4}

For

u^{n} < a

, if

n

is even then

- \frac{29}{4} < x + \frac{5}{2} < \frac{29}{4}

a < u < b

then

a < u and u < b

- \frac{29}{4} < x + \frac{5}{2} and x + \frac{5}{2} < \frac{29}{4}

- \frac{29}{4} < x + \frac{5}{2} : x > \frac{- 29 - 5}{2}

x + \frac{5}{2} < \frac{29}{4} : x < \frac{29 - 5}{2}

Combine the intervals

x > \frac{- 29 - 5}{2} and x < \frac{29 - 5}{2}

Merge Overlapping Intervals

\frac{- 29 - 5}{2} < x < \frac{29 - 5}{2}

\frac{6}{- 8}

\frac{441 u ^{3}}{12 v ^{8}}

\frac{3}{2} x + 1 = 2

(3 x^{\frac{5}{6}}) (8 x^{\frac{2}{3}})

\frac{25 !}{10 ! \cdot ( 25 - 10 ) !}

What is x^2+5x-1<0 ?
The solution to x^2+5x-1<0 is (-sqrt(29)-5)/2 <x<(sqrt(29)-5)/2