The solution set of the inequation $\sqrt{x^2+6 x+5}>(8-x)$ is

The solution set of the inequation $\sqrt{x^2+6 x+5}>(8-x)$ is
  1. $(8, \infty)$
  2. $\left(\frac{59}{22}, 8\right]$
  3. $\left(\frac{59}{22}, \infty\right)$
  4. $(-1, \infty)$

Solution

$\sqrt{x^2+6 x+5}>(8-x)$ On squaring both side, $\begin{gathered} \left(\sqrt{x^2+6 x+5}\right)^2>(8-x)^2 \\ x^2+6 x+5>64-16 x+x^2 \\ 6 x+16 x>64-5 \\ 22 x>59 \\ x>\frac{59}{22} \\ x \in\left(\frac{59}{22}, \infty\right) \end{gathered}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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