integer roots...

are there no other val. of a for which a² - 24a is a perfect square. How can we say this?

\hspace{-16}$If $\mathbf{x^2+ax+6a=0}$ has Integer Roots. Then $\mathbf{a\in \mathbb{Z}}$\\\\ So $\mathbf{a=-\frac{x^2}{x+6}=-x+6-\frac{36}{x+6}}$\\\\ Now $\mathbf{a\in \mathbb{Z}}$, If $\mathbf{\frac{36}{x+6}\in \mathbb{Z}}$\\\\ Which is Possible only When\\\\ $\mathbf{x+6\in \{\pm 1,\pm 2\;,\pm 3\;,\pm4\;,\pm 6\;,\pm 9\;,\pm 12\;,\pm 18\;,\pm 36 \}}$\\\\ So $\mathbf{x\in \{(-5,-7)\;,(-4,-8)\;,(-3,-9)\;,(-2,-10)\}}$\\\\ and $\mathbf{x\in \{(0,-12)\;,(3,-15)\;,(6,-18)\;,(12,-24)\;, (30,-42)\}}$\\\\ So $\mathbf{a\in \{0,\pm 1\;,\pm 3\;,\pm 6,\pm 8\;,\pm 18\;,\pm 24\;,\pm 25\;,\pm 27\;,\pm 32\;,\pm 49\}}$\\\\ and $\mathbf{D=a^2-24a\geq 0 \Leftrightarrow a\leq 0\; \cup \geq 24}$

Oh my God.

You are Right Rahul. (x+6 = ±36)

actually Writting in Latex editor is very Complicated for Me.

Thats why Mistake is Persent

So Sorry for that.

Thnaks Rahul for Pointing.