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Find the value of $x$ such that \[{x^2} + 2x,2x + 3\] and \[{x^2} + 3x + 8\] are the length of the sides of a triangle.

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Hint: A triangle is a polygon having three sides and three vertices. In geometry, it is a closed two-dimensional shape having three straight sides. In this question, a very basic property of the triangle is used: that is the sum of the two sides of the triangle is more than the third side.

Complete step by step solution:Check whether these three sides form a triangle or not, to check we use the Triangle inequality theorem which says that the sum of any two side length of a triangle is always greater than its third side length.
By using this property of triangle we can write
\[\left( {{x^2} + 2x} \right) + \left( {2x + 3} \right) > {x^2} + 3x + 8 - - - - (i)\]
Now solve for \[x\] we can write:
\[ \left( {{x^2} + 2x} \right) + \left( {2x + 3} \right) > {x^2} + 3x + 8 \\
  {x^2} + 4x + 3 > {x^2} + 3x + 8 \\
  4x - 3x > 8 - 3 \\
  x > 5 - - - - (ii) \\ \]
Now for the other side
\[\left( {{x^2} + 2x} \right) + {x^2} + 3x + 8 > \left( {2x + 3} \right) - - - - (iii)\]
Solving for \[x\]:
\[ \left( {{x^2} + 2x} \right) + {x^2} + 3x + 8 > \left( {2x + 3} \right) \\
  2{x^2} + 5x + 8 > 2x + 3 \\
  2{x^2} + 5x - 2x > 3 - 8 \\
  2{x^2} + 3x + 5 > 0 \\
  x = - 0.75 \pm 1.39i \\ \]
The above equation will have the complex values of \[x\] and so can be neglected.
Now check for the third side,
\[\left( {2x + 3} \right) + {x^2} + 3x + 8 > \left( {{x^2} + 2x} \right) - - - - (iv)\]
solving for the value of \[x\]:
\[ \left( {2x + 3} \right) + {x^2} + 3x + 8 > \left( {{x^2} + 2x} \right) \\
  {x^2} + 5x + 11 > {x^2} + 2x \\
  3x > - 11 \\
  x > \dfrac{{ - 11}}{3} \\ \]
The values obtained \[x\] from the three equations (i), (ii), and (iii) we can say that the length of the side of a triangle can never be in negative or in complex form, hence the value of \[x\] the side of the triangle is \[x > 5\].

Note: If the equality theorem of a triangle does not satisfy then that polygon can never form a triangle. There are three basic types of triangles, the first one is the isosceles triangle, these triangles have any two sides equal, the second one is the equilateral triangle, these triangles have all sides equal and the third one is the scalene triangle, these triangles all sides are unequal. There is one more type of triangle known as the right-angled triangle in which one of the three angles is at 90 degrees.