The real numbers x satisfying frac{sqrt{x+5}} | vedclass

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(D) Given the inequality $\frac{\sqrt{x+5}}{1-x} > 1$. For the expression to be defined,we must have $x+5 \ge 0 \Rightarrow x \ge -5$ and $1-x 
eq 0

Vedclass · Accepted Answer

$-1 < x < 1$

Vedclass · Answer

(D) Given the inequality $\frac{\sqrt{x+5}}{1-x} > 1$. For the expression to be defined,we must have $x+5 \ge 0 \Rightarrow x \ge -5$ and $1-x eq 0 \Rightarrow x eq 1$. Case $1$: If $1-x > 0$,i.e.,$x 1-x$. Since $x \ge -5$,$x+5$ is non-negative. If $1-x 0$. Squaring both sides: $x+5 > (1-x)^2 = 1 - 2x + x^2$. $x^2 - 3x - 4 $(x-4)(x+1) This implies $-1 Combining with $x Case $2$: If $1-x 1$,then $\sqrt{x+5} Since $\sqrt{x+5} \ge 0$ and $1-x Thus,the solution is $-1 < x < 1$.

The real numbers $x$ satisfying $\frac{\sqrt{x+5}}{1-x} > 1$ are precisely those which satisfy

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