Find the real value of x such that x2+2x,2x+3 and x2+3x+8 are lengths of the sides of a triangle.
x2+2x,2x+3, & x2+3x+8
are the sides of a triangle
then sum of any two side will always be greater than the third side
(x2+2x)+(2x+3)>x2+3x+8
⇒x>5 ………(1)
(x2+2x)+(x2+3x+8)>2x+3
2x2+3x+5>0
⇒x∈R (∴a>0,D<0)
(2x+3)+(x2+3x+8)>x2+2x
3x+11>0
x>−113
So these will be sides of a triangle for x>5.