If x is real and satisfies x + 2 sqrt {x | vedclass

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Question

(b) Given , $x + 2 > \,\sqrt {x + 4} $$ \Rightarrow {(x + 2)^2}\,\, > (x + 4)$

$ \Rightarrow {x^2} + 4x + 4 > x + 4$ $ \Rightarrow {x^2} +

Vedclass · Accepted Answer

$x > 0$

Vedclass · Answer

(b) Given , $x + 2 > \,\sqrt {x + 4} $$ \Rightarrow {(x + 2)^2}\,\, > (x + 4)$

$ \Rightarrow {x^2} + 4x + 4 > x + 4$ $ \Rightarrow {x^2} + 3x > 0$

$ \Rightarrow x(x + 3) > 0$ ==> $x < -3$ or $x > 0$ $ \Rightarrow \,x > 0$.

If $x$ is real and satisfies $x + 2 > \sqrt {x + 4} ,$ then

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