Find all \(a\in\mathbb{R}\) such that \((a+2)x^2-(a+1)x+a=0\) has at least one integer solution
@oldrin.bataku
would this pertain to the discriminant of the quadratic equation? \[b^2-4ac\ge0\]???
For both the roots to be integer in above equation (a+2) has to be 1 which can be obtained by diving both sides by (a+2) and then the discriminant has to be perfect square.. But i don't know what am i supposed to do when at least one root has to be an integer..
if at least 1 root is an integer would require that there is a non-imaginary root to begin with ... so the discriminant greater or equal to 0 allows us to define it better; so \[(a+1)^2-4(a+2)(a)\ge0\] \[a^2+2a+1-4a^2+8a\ge0\] \[-3a^2+10a+1\ge0\] we know have another quadratic in "a" to determine ...
since the first term is negative, this will most likely be a domain between the zeros
using the quadratic formula for "a"; this is equal to zero when\[a=\frac{-10\pm\sqrt{10^2-4(-3)(1)}}{2(-3)}\]
notice that we do not need both roots to be an integer; only one of them would have to be an integer (at least one means: 1 or more)
ensuring that the discriminant of the first setup is a perfect square would not be sufficient to give a value of a that is an integer to start with ... but then it might if the problem was constructed that way
\[(a+2)x^2-(a+1)x+a=0\] \[(a+2)~\left(x^2-\frac{a+1}{a+2}x+\frac{a}{a+2}\right)=0\] \[(a+2)~\left(x^2-\frac{a+1}{a+2}x\pm\left(\frac{a+1}{2(a+2)}\right)^2+\frac{a}{a+2}\right)=0\] \[(a+2)~\left(x-\frac{a+1}{2(a+2)}\right)^2+a-a+1=0\] \[\left(x-\frac{a+1}{2(a+2)}\right)^2=\frac{-1}{a+2}\] \[x-\frac{a+1}{2(a+2)}=\pm\sqrt{\frac{-1}{a+2}}\] \[x=\frac{a+1}{2(a+2)}\pm\sqrt{\frac{-1}{a+2}}\] \[x=\frac{a+1\pm\sqrt{-4(a+2)}} {2(a+2)}\] \[-4a-8\ge0~:~a\le-2\] maybe
Discriminant=\(1-3a^2-6a\) which is real only when a=0, -1 and -2 and coincidently, at those values, the equation gives integer solutions.. Solved!
:) yay
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