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The set of all real numbers \mathrm{x} such that \mathrm{x^2+3 x+8, x^2+2 x \: and \: 2 x+3} are the lengths of the sides of a triangle is given by
 

Option: 1

\mathrm{x>5}

 


Option: 2

\mathrm{x<5}
 


Option: 3

\mathrm{x>2}
 


Option: 4

\mathrm{x \leq 5}


Answers (1)

best_answer

As the sum of the lengths of two sides of a triangle is greater than the length of the third side,
\mathrm{ \left(x^2+3 x+8\right)+\left(x^2+2 x\right)>2 x+3 }.................(i)
\mathrm{ \left(x^2+3 x+8\right)+(2 x+3)>x^2+2 x }.................(ii)
\mathrm{ \left(x^2+2 x\right)+(2 x+3)>x^2+3 x+8 }.................(iii)

From (i), \mathrm{x^2+\frac{3}{2} x+\frac{5}{2}>0 \Rightarrow\left(x+\frac{3}{4}\right)^2+\frac{31}{16}>0}is true for \mathrm{x \in R}From(ii), \mathrm{\quad x>-\frac{11}{3}}

and from (iii), \mathrm{x>5}

These inequalities are satisfied when \mathrm{x>5}.

Hence option 1 is correct.

Posted by

Gunjita

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