The solution to x^2+3x+1<0 is (-sqrt(5)-3)/2 <x<(sqrt(5)-3)/2

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x^2+3x+1<0

Solution

x^{2} + 3 x + 1 < 0

\frac{- 5 - 3}{2} < x < \frac{5 - 3}{2}

Interval Notation

(\frac{- 5 - 3}{2}, \frac{5 - 3}{2})

Decimal Notation

- 2.61803 \dots < x < - 0.38196 \dots

Solution steps

x^{2} + 3 x + 1 < 0

Complete the square

x^{2} + 3 x + 1 : (x + \frac{3}{2})^{2} - \frac{5}{4}

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

Move

\frac{5}{4}

to the right side

(x + \frac{3}{2})^{2} < \frac{5}{4}

For

u^{n} < a

, if

n

is even then

- \frac{5}{4} < x + \frac{3}{2} < \frac{5}{4}

a < u < b

then

a < u and u < b

- \frac{5}{4} < x + \frac{3}{2} and x + \frac{3}{2} < \frac{5}{4}

- \frac{5}{4} < x + \frac{3}{2} : x > \frac{- 5 - 3}{2}

x + \frac{3}{2} < \frac{5}{4} : x < \frac{5 - 3}{2}

Combine the intervals

x > \frac{- 5 - 3}{2} and x < \frac{5 - 3}{2}

Merge Overlapping Intervals

\frac{- 5 - 3}{2} < x < \frac{5 - 3}{2}

121 x^{2} - 22 x + 1 < 0

(x^{2} + 3 x + 13) > 0

(x + 1)^{2} \cdot (x - 1)^{3} \cdot (x - 5)^{7} \geq 0

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

\frac{x}{2 x} < 9

What is x^2+3x+1<0 ?
The solution to x^2+3x+1<0 is (-sqrt(5)-3)/2 <x<(sqrt(5)-3)/2