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Properties of Determinants

Find the real...

Question

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

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Solution

$x^{2} + 2 x, 2 x + 3$ , & $x^{2} + 3 x + 8$
are the sides of a triangle
then sum of any two side will always be greater than the third side
$(x^{2} + 2 x) + (2 x + 3) > x^{2} + 3 x + 8$
$\Rightarrow x > 5$ ……… $(1)$
$(x^{2} + 2 x) + (x^{2} + 3 x + 8) > 2 x + 3$
$2 x^{2} + 3 x + 5 > 0$
$\Rightarrow x \in R$ $(∴ a > 0, D < 0)$
$(2 x + 3) + (x^{2} + 3 x + 8) > x^{2} + 2 x$
$3 x + 11 > 0$
$x > \frac{- 11}{3}$
So these will be sides of a triangle for $x > 5$ .

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