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, is-

• A. $(5,\infty )$
• B. $(3,\infty )$
• C. $(2,\infty )$
• D. $(0,\infty )$

Answer 1

The correct option is A $(5,\infty )$

Given, $a^2+2a,2a+3$ and $a^2+3a+8$ are the sides of a triangle
We must have
$a^2+2a>0$ (since these are sides)
$\Rightarrow a(a+2)>0$
$\Rightarrow a<-2$ or $a>0$
Also, $2a+3>0$ (since these are sides)
$\Rightarrow a>-\frac{3}{2}$
Also, $a^2+3a+8>0$ (since these are sides)
$a^2+3a+8>0\forall a\in R$
We also have,
$a^2+2a+2a+3>a^2+3a+8$ (Sum of two sides must be greater than the third side)
$\Rightarrow a>5$
$a^2+2a+a^2+3a+8>2a+3$ (Sum of two sides must be greater than the third side)
$\Rightarrow 2a^2+3a+5>0\forall a\in R$
$a^2+3a+8+2a+3>a^2+2a$ (Sum of two sides must be greater than the third).
$\Rightarrow a>-1$
So, the common solution is $(5,\infty )$