The solution to x^2+8x+5<0 is -sqrt(11)-4<x<sqrt(11)-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+8x+5<0

Solution

x^{2} + 8 x + 5 < 0

- 11 - 4 < x < 11 - 4

Interval Notation

(- 11 - 4, 11 - 4)

Decimal Notation

- 7.31662 \dots < x < - 0.68337 \dots

Solution steps

x^{2} + 8 x + 5 < 0

Complete the square

x^{2} + 8 x + 5 : (x + 4)^{2} - 11

(x + 4)^{2} - 11 < 0

Move

11

to the right side

(x + 4)^{2} < 11

For

u^{n} < a

, if

n

is even then

- 11 < x + 4 < 11

a < u < b

then

a < u and u < b

- 11 < x + 4 and x + 4 < 11

- 11 < x + 4 : x > - 11 - 4

x + 4 < 11 : x < 11 - 4

Combine the intervals

x > - 11 - 4 and x < 11 - 4

Merge Overlapping Intervals

- 11 - 4 < x < 11 - 4

a^{6} b^{7} - a^{4} b^{3} + a^{3} b^{5} - a^{3} b^{3}

4^{∣ x + 8 ∣} = 64

2 a c - 5 b d - 2 a + 2 a d + 5 b - 5 b c

(32 x^{30} y^{- 20})^{- \frac{3}{5}}

y - 9 = 0