1.

|x2 + x| - 5 < 0.

Answer»

\(|x^2 +x |-5 < 0\)

⇒ \(|x (x + 1)| - 5 < 0\)

Case I: \(x \le 0\)

Then

\(x(x + 1)> 0\)

⇒ \(|x (x + 1)| = x(x +1)\)

⇒ \(|x^2 + x| - 5 < 0\)

⇒ \(x^2 + x - 5 < 0\)

⇒ \(\frac{-1-\sqrt{21}}{2}< x< \frac{-1 + \sqrt{21}}{2}\)

But \(x\le -1\)

\(\therefore x \in \left(\frac{-1-\sqrt{21}}{2}, -1\right]\)    .....(1)

Case II: \(- 1 < x \le 0\)

Then

\(x(x + 1)< 0\)

⇒ \(|x^2 + x| = -(x^2 +1)\)

⇒ \(|x^2 + x|-5 < 0\)

⇒ \(- (x^2 + x) - 5 < 0\)

⇒ \(x^2 + x + 5 > 0\)

⇒ \(x\in R\)

Hence, \(x \in ( -1 , 0]\)     .....(2)

Case III: x > 0

Then

 \(x(x + 1)> 0\)

⇒ \(|x^2 + x| = x^2 +x\)

\(\therefore |x^2 + x| - 5 < 0\)

⇒ \(x^2 +x - 5 < 0\)

⇒ \(\frac{-1 - \sqrt{21}}{2} < x < \frac{-1 + \sqrt{21}}{2}\)

But \(x > 0\)

\(\therefore x \in \left(0,\frac{-1+\sqrt{21}}{2}\right)\)    ....(3)

From (1), (2) & (3), we get 

\(x \in \left(\frac{-1-\sqrt{21}}{2},\frac{-1 + \sqrt{21}}{2}\right)\)


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