\(|x^2 +x |-5 < 0\)
⇒ \(|x (x + 1)| - 5 < 0\)
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
\(x \in \left(\frac{-1-\sqrt{21}}{2},\frac{-1 + \sqrt{21}}{2}\right)\)
