The set of all real numbers x for which {x^2} | vedclass

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Question

(b) Case I: When $x + 2 \ge 0$ i.e. $x \ge - 2,$

Then given inequality becomes

${x^2} - (x + 2) + x > 0$ ==> ${x^2} - 2 > 0 \Rightarrow

Vedclass · Accepted Answer

$( - \infty ,\,\, - \sqrt 2 )\, \cup (\sqrt 2 ,\,\infty )$

Vedclass · Answer

(b) Case I: When $x + 2 \ge 0$ i.e. $x \ge - 2,$

Then given inequality becomes

${x^2} - (x + 2) + x > 0$ ==> ${x^2} - 2 > 0 \Rightarrow \,\,|x|\, > \sqrt 2 $

==> $x < - \sqrt 2 $ or $x > \sqrt 2 $

As $x \ge - 2,$ therefore, in this case the part of the solution set is $[ - 2, - \sqrt 2 ) \cup (\sqrt 2 ,\infty ).$

Case II: When $x + 2 \le 0$ i.e. $x \le - 2,$

Then given inequality becomes ${x^2} + (x + 2) + x > 0$

$ \Rightarrow {x^2} + 2x + 2 > 0$ $ \Rightarrow {(x + 1)^2} + 1 > 0,$ which is true for all real $x$

Hence, the part of the solution set in this case is $( - \infty , - 2]$.

Combining the two cases, the solution set is

$( - \infty , - 2) \cup ([ - 2, - \sqrt 2 ] \cup (\sqrt 2 ,\infty )$ $ = ( - \infty , - \sqrt 2 ) \cup (\sqrt 2 ,\infty ).$

The set of all real numbers $x$ for which ${x^2} - |x + 2| + x > 0,$ is

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