Solve the inequality and express in interval notation x^2 + 8x + 5 < 0

Hello from MrBillDoesMath!

x > -4 - sqrt(11) and x < -4 + sqrt(11)

Discussion:

Let's complete the square by adding and subtracting the value 16 from both sides

x^2 + 8x + 5 < 0   =>

(x^2 + 8x + 16) + 5 < 0 + 16 =>      complete the square

(x + 4) ^2 + 5  < 16 =>                    subtract 5 from both sides

(x+4)^2 < 16 - 5 = 11  =>                  take square root of both sides

(x+4) > -sqrt(11)  and ( x+4 ) < sqrt(11)  =>

x > -4 - sqrt(11) and x < -4 + sqrt(11)

Thank you,

MrB

its not b

step-by-step explanation:

step-by-step explanation:

the true ones are a b and d