Question

Find the real value of $x$ such that $x^2+2x,2x+3$ and $x^2+3x+8$ are lengths of the sides of a triangle.

Answer 1

$x^2+2x,2x+3$, & $x^2+3x+8$

are the sides of a triangle

then sum of any two side will always be greater than the third side

$(x^2+2x)+(2x+3)>x^2+3x+8$

$\Rightarrow x>5$ ………$\left(1\right)$

$(x^2+2x)+(x^2+3x+8)>2x+3$

$2x^2+3x+5>0$

$\Rightarrow x\in R$ $(\therefore a>0,D<0)$

$(2x+3)+(x^2+3x+8)>x^2+2x$

$3x+11>0$

$x>\frac{-11}{3}$

So these will be sides of a triangle for $x>5$.