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

Answer:

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

Step-by-step explanation:

Given Quadratic inequality ,

$4(x+2)^2$

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

consider,

$4(x+2)^2$

transpose 4 to RHS

$(x+2)^2$

$(x+2)^2$

Square root both side,

$\sqrt{(x+2)^2}$

$x+2$

transpose 2 to RHS

$x$

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 }