Question

X-3< √(x^2+4x-5) .Please solve this inequality.​

Answer 1

Given Inequality

$\sf{x - 3} < \sqrt{x {}^{2} + 4x + 5}$

Squaring on both sides,we get:

$\implies \: \sf{(x - 3) {}^{2} < x {}^{2} + 4x + 5 } \\ \\ \implies \: \sf{ \cancel{x {}^{2}} + 9 - 6x < \cancel{x {}^{2}} + 4x + 5} \\ \\ \implies \: \sf{10x > 9 - 5} \\ \\ \implies \: \sf{10x > 4} \\ \\ \implies \: \huge{ \sf{x > \frac{2}{5} }}$