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How to find the discriminant of this equation:


Because, if I'm correct, that becomes: $x^2+32+12ax+a^2x^2$, and I have no clue how to continue..

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up vote 2 down vote accepted

The equation can arranged as $x^2+x(a-4)+2=0$

So, the discriminant is $(a-4)^2-4\cdot1\cdot 2=a^2-8a+8$

With the edited question,the equation can arranged as $x^2(1+a^2)+x(-4+12a)+32=0$

So, the discriminant is $(12a-4)^2-4\cdot(a^2+1)\cdot32=16a^2-96a-112$

SO, $$x=\frac{-(12a-4)\pm\sqrt{16a^2-96a-112}}{2\cdot (1+a^2)}=\frac{2-6a\pm\sqrt{4a^2-24a-28}}{a^2+1}$$

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Shouldn't it be $a^2-8a-112$? Note that you used 2 instead of 32, but even then you did something wrong I believe $(a-4)^2 = a^2 -8a +16$ – JohnPhteven Oct 23 '12 at 12:17
@ZafarS, rectified my answer. But I'm not sure where $32$ came from? – lab bhattacharjee Oct 23 '12 at 12:19
$(ax+6)^2 = a^2x^2 + 12ax+36$, and there is a $4$ on the RHS. – JohnPhteven Oct 23 '12 at 12:20
@ZafarS, 1st of all, we need to express as $ax^2+bx+c=0,$ then the discriminant will be $b^2-4ac;a,b$ and $c$ are independent of $x$. – lab bhattacharjee Oct 23 '12 at 12:24
I forgot to put a square in the original question, will you now please do it again and solve the quadratic equation, I don't know how to in this case – JohnPhteven Oct 23 '12 at 12:29

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