Question

Find the solution of the inequality $(x-3)<\sqrt{x^2+4x-5}$

• A. $(-\infty ,-5]\cup [1,\infty )$
• B. $(-\infty ,-5]$
• C. $(\frac{7}{5},\infty )$
• D. $(-\infty ,1]\cup [5,\infty )$

Answer 1

The correct option is C $(\frac{7}{5},\infty )$

$(x-3)<\sqrt{x^2+4x-5}$
$\Rightarrow (x-3{)}^2<x^2+4x-5$
$\Rightarrow -10x<-14$
$\Rightarrow x>\frac{7}{5}$ .....(1)
From the given inequality, it follows that
$x^2+4x-5\ge 0$
$\Rightarrow (x+5)(x-1)\ge 0$
$\Rightarrow x\in (-\infty ,-5]\cup [1,\infty )$ .....(2)
From (1) and (2), it follows that
$\Rightarrow x\in (\frac{7}{5},\infty )$