Question

The set of all real numbers $a$ such that $a^2+2a,2a+3$ and $a^2+3a+8$ are the sides of a triangle $(a,\infty )$ then a is

Answer 1

If $a^2+2a,2a+3$ and $a^2+3a+8$ are the sides of a triangle, then sum of any two sides is greater than the third side.
Let $x=a^2+2a;y=2a+3;z=a^2+3a+8$
Then, $x+y>z$
$\Rightarrow a^2+2a+2a+3>a^2+3a+8$
$\Rightarrow a^2+4a+3>a^2+3a+8$
$\Rightarrow a>5$ ...............$\left(1\right)$
$y+z>x$
$\Rightarrow 2a+3+a^2+3a+8>a^2+2a$
$\Rightarrow a^2+5a+11-a^2-2a>0$
$\Rightarrow 3a>-11$
$\Rightarrow a>\frac{-11}{3}$ ...............$\left(2\right)$
$z+x>y$
$\Rightarrow a^2+3a+8+a^2+2a>2a+3$
$\Rightarrow 2a^2+5a+8-2a-3>0$
$\Rightarrow 2a^2+3a+5>0$
Here, coefficient of $a^2>0$ and $D=9-40=$negative.
$\therefore$ It is true for all values of $a$
$\therefore$ Identity combining eqns$\left(1\right)$ and $\left(2\right)$, we get
$a>5$ or $a\in (5,\infty )$