ridgers wrote: The given parabola $y=ax^2+bc+c$ doesn't intersect the $X$-axis and passes from the points $A(-2,1)$ and $B(2,9)$. Find all the possible values of the $x$ coordinates of the vertex of this parabola. No such parabola exists since the function $f(x)=ax^2+bc+c$ is even and so $f(2)=f(-2)$ and we cant have $f(2)=9$ and $f(-2)=1$

ridgers wrote: The given parabola $y=ax^2+bc+c $ doesn't intersect the $X$-axis and passes from the points $A(-2,1)$ and $B(2,9)$. Find all the possible values of the $x$ coordinates of the vertex of this parabola. Check! $y=ax^2+bc+c \iff y-c(b+1)=Y=ax^2; (P)$ then $f(-x)=f(x)$, $x=2$???

ridgers wrote: The given parabola $y=ax^2+bx+c$ doesn't intersect the $X$-axis and passes from the points $A(-2,1)$ and $B(2,9)$. Find all the possible values of the $x$ coordinates of the vertex of this parabola. From $4a-2b+c=1$ and $4a+2b+c=9$, we get $b=2$ and $c=5-4a$ From the condition "no real roots", we get $b^2-4ac=4-4a(5-4a)=16a^2-20a+4<0$ and so $a\in(\frac 14,1)$ And so $\boxed{-\frac b{2a}\in(-4,-1)}$

@ridgers: Are you sure this is the subject of mathematics in Albanian IMO 2011 TST Simply because it is very basic!

tuan119 wrote: @ridgers: Are you sure this is the subject of mathematics in Albanian IMO 2011 TST Simply because it is very basic! Yes I am pretty sure cause I was in TST! Albanian level is not so high!

easy one!)

A geometrical argument shows that there are two extremal position for the parabola, one with vertex that has $x$-coordinate between $A$ and $B$, the other with the vertex to the left of $A$, and by continuity all positions between the two are attained. In both limiting cases it is tangent to the $x$ axis, so $b^2=4ac$. Imposing the parabola to pass through $A$ and $B$ we immediately get $b=2$; note that the $x$-coordinate of the vertex is $-\frac b {2a}$. We get a system $4a-4 \sqrt{ac}+c=1, 4a+4 \sqrt {ac}+c=9$ which gives as solutions $(a, c)=(1, 1), (\frac 1 4 , 4)$, corresponding to an $x$-coordinate of the vertex respectively of $-1$ and $-4$, so the desired set of values we are searching for is the interval $(-4, -1)$.

Solution already posted in this thread, but it should be fine. Substitution of $x = -2, \ y = 1$ and $x = 2, \ y = 9$ yields the equations $1 = 4a - 2b + c$ and $9 = 4a + 2b + c$. Eliminating the terms $4a$ and $c$ yields $b = 2$, and eliminating the $2b$ and $-2b$ terms yields $c = 5 - 4a$ ($\spadesuit$). The discriminant needs to be less than $0$. Using ($\spadesuit$), we write $$b^2 - 4ac < 0 \Longleftrightarrow ac > 1 \Longleftrightarrow 4a^2 - 5a + 1 < 0 \Longleftrightarrow (4a - 1)(a - 1) < 0.$$It is clear that $a \in (\frac{1}{4}, 1)$, and thus $\boxed{-\frac{b}{2a} = -\frac{1}{a} \in (-4, -1)}$, as desired.