The set of all real numbers x for which \( \Large x^{2}-|x+2|+x>0 \) is:
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A) \( \Large \left(-\infty, -2\right) \ \left(2, \infty\right) \) |
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B) \( \Large \left(-\infty, -\sqrt{2}\right) \ \left(\sqrt{2}, \infty\right) \) |
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C) \( \Large \left(-\infty, -1\right) \ \left(1, \infty\right) \) |
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D) \( \Large \left(\sqrt{2}, \infty\right) \) |
Correct Answer:
| B) \( \Large \left(-\infty, -\sqrt{2}\right) \ \left(\sqrt{2}, \infty\right) \) |
Given \( \Large x^{2}-|x+2|+x>0 \)
we know that\( \Large |x+2|=\pm \left(x+2\right) \)
If \( \Large |x+2|=-x+2x\ge -2 \)
Equation (i) becomes \( \Large \left(when\ x \ge - 2 \right) \)
\( \Large x^{2}-x-2+x>0 \)
\( \Large x^{2}-2 > 0 \)
=> \( \Large x < -\sqrt{2}\ or\ x > \sqrt{2} \)
\( \Large x \epsilon \left[ -2, -\sqrt{2} \right] \cup \left(\sqrt{2}, \infty\right) \)
If therefore, equation (i) becomes
\( \Large x^{2}+x+2+x>0\ when\ x \le -2 \)
\( \Large x^{2}+2x+2 > 0\ when\ x \le -2 \)
\( \Large \left(x+1\right)^{2}+1 > 0\ when x \le -2 \)
which is true for all x.
\( \Large x < -2\ or\ x \left(-\infty,\ -2\right) \)
from equations (ii) and (iii)
\( \Large x \epsilon \left(-\infty,\ -\sqrt{2} \right) \cup \left(\sqrt{2},\ \infty\right) \)
Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations
