What value of $k$ makes $x^{2} + kx + 36$ a perfect square (with $k>0$)?

What value of $k$ makes $x^{2} + kx + 36$ a perfect square (with $k>0$)?
  1. $12$
  2. $6$
  3. $18$
  4. $36$

Solution

For $(x+a)^{2}=x^{2}+2ax+a^{2}$, here $a^{2}=36 \Rightarrow a=6$, so $k = 2a = 12$.

Asked in: IMO

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