Question

If $\sqrt{x^2+x-12}<6-x$, then

• A. $x\in (-\infty ,-4]\cup [3,\infty )$
• B. $x\in [3,\infty )$
• C. $x\in (-\infty ,-4]$
• D. $x\in (-\infty ,-4]\cup [3,\frac{48}{13})$

Answer 1

The correct option is D $x\in (-\infty ,-4]\cup [3,\frac{48}{13})$

Given $\sqrt{x^2+x-12}<6-x$ to be defined
$x^2+x-12\ge 0$
$(x+4)(x-3)\ge 0$
$\Rightarrow x\in (-\infty ,-4]\cup [3,\infty )\cdots \left(1\right)$

Now
$\sqrt{x^2+x-12}<6-x$ squaring both sides
$\Rightarrow x^2+x-12<36+x^2-12x\Rightarrow 13x<48\Rightarrow x<\frac{48}{13}\cdots \left(2\right)$

from $\left(1\right)$ and $\left(2\right)$ we get
$x\in (-\infty ,-4]\cup [3,\frac{48}{13})$