The solution to x^2+8x+3<0 is -sqrt(13)-4<x<sqrt(13)-4

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x^2+8x+3<0

Solution

x^{2} + 8 x + 3 < 0

- 13 - 4 < x < 13 - 4

Interval Notation

(- 13 - 4, 13 - 4)

Decimal Notation

- 7.60555 \dots < x < - 0.39444 \dots

Solution steps

x^{2} + 8 x + 3 < 0

Complete the square

x^{2} + 8 x + 3 : (x + 4)^{2} - 13

(x + 4)^{2} - 13 < 0

Move

13

to the right side

(x + 4)^{2} < 13

For

u^{n} < a

, if

n

is even then

- 13 < x + 4 < 13

a < u < b

then

a < u and u < b

- 13 < x + 4 and x + 4 < 13

- 13 < x + 4 : x > - 13 - 4

x + 4 < 13 : x < 13 - 4

Combine the intervals

x > - 13 - 4 and x < 13 - 4

Merge Overlapping Intervals

- 13 - 4 < x < 13 - 4

(2 x + 3 y + a)^{2} - (x - y + a)^{2}

0 < 6 x - x^{2} - 5

\frac{5}{- 5 - 3 3}

- 17 + \frac{n}{5} = 33