Question

What is the solution set of the quadratic inequality 4(x+2)^2<0

Answer 1

`Answer: \\ 4 {(x + 2)}^(2) < 0 \\ \Leftrightarrow ( {x + 2})^(2) < 0 \: which \: is \: wrong \: because \: {(x + 2)}^(2) \geqslant 0 \\ \Rightarrow x \in \emptyset`

Answer 2

Solution set of the quadratic inequality is { x : x ∈ R and x < -2 }

Explanation:

Given Quadratic inequality ,

`4(x+2)^2<0`

We have to find solution set of the given quadratic inequality.

consider,

`4(x+2)^2<0`

transpose 4 to RHS

`(x+2)^2<(0)/(4)`

`(x+2)^2<0`

Square root both side,

`√((x+2)^2)<√(0)`

`x+2<0`

transpose 2 to RHS

`x<0-2`

x < -2

Solution set of the quadratic inequality = { x : x ∈ R and x < -2 }

Therefore, Solution set of the quadratic inequality is { x : x ∈ R and x < -2 }