If x is real and satisfies x + 2 > √ {x + 4} , then

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4-2.Quadratic Equations and Inequations

medium

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

A

$x < - 2$

B

$x > 0$

C

$ - 3 < x < 0$

D

$ - 3 < x < 4$

Solution

(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$.

Standard 11

Mathematics

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