### Instruction: A parabola with equation $y=ax^2+bx+c$ has a vertical line of symmetry at $x=1$ and goes through the two points $(-1,3)$ and $(2,-2)$. The quadratic $ax^2 + bx +c$  has two real roots. The greater root is $\sqrt{n}+1$. What is $n$?### Response: I know that a parab68ola with a vertical line of symmetry at $x=1$ must have its vertex at $x=1$, so the equation of the parabola must be of the form $y=a(x-1)^2+k$, where $a$ and $k$ are constants.
I can use the two given points to set up a system of equations to find $a$ and $k$.
If I plug in $x=-1$ and $y=3$, I get $3=a(-1-1)^2+k$, or $3=4a+k$.
If I plug in $x=2$ and $y=-2$, I get $-2=a(2-1)^2+k$, or $-2=a+k$.
Subtracting the second equation from the first, I get $5=3a$, or $a=\frac{5}{3}$.
Plugging this value into the second equation, I get $-2=\frac{5}{3}+k$, or $k=-\frac{11}{3}$.
So the equation of the parabola is $y=\frac{5}{3}(x-1)^2-\frac{11}{3}$.
To find the roots of the quadratic, I need to set $y=0$ and solve for $x$.
So I have $0=\frac{5}{3}(x-1)^2-\frac{11}{3}$.
Multiplying both sides by $3$, I get $0=5(x-1)^2-11$.
Adding $11$ to both sides, I get $11=5(x-1)^2$.
Dividing both sides by $5$, I get $\frac{11}{5}=(x-1)^2$.
Taking the square root of both sides, I get $\pm\sqrt{\frac{11}{5}}=x-1$.
Adding $1$ to both sides, I get $x=1\pm\sqrt{\frac{11}{5}}$.
These are the two roots of the quadratic, and the greater one is $1+\sqrt{\frac{11}{5}}$.
I need to find $n$ such that $1+\sqrt{\frac{11}{5}}=\sqrt{n}+1$.
Subtracting $1$ from both sides, I get $\sqrt{\frac{11}{5}}=\sqrt{n}$.
Squaring both sides, I get $\frac{11}{5}=n$.
This is the value of $n$ I was looking for.