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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If for a posiive integer $n$ , the quadratic equation, $x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .\;.\;.\; + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$ has two consecutive integral solutions, then $n$ is equal to:

If the roots of ${x^2} + x + a = 0$exceed $a$, then

If $\alpha ,\beta $ and $\gamma $ are the roots of ${x^3} + px + q = 0$, then the value of ${\alpha ^3} + {\beta ^3} + {\gamma ^3}$ is equal to

Suppose $m, n$ are positive integers such that $6^m+2^{m+n} \cdot 3^w+2^n=332$. The value of the expression $m^2+m n+n^2$ is

Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be