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I need to solve a quadratic equation (actually I need to explain it to my kid), but I get stuck in the middle and would be grateful, for any pointers into the right direction.

$y=ax^2+bx-1$

with two points on its parabola: $P_1=(-2,1)$ and $P_2=(3,1)$. I need to find $a$ and $b$.

Inserting the coordinates, I get \begin{align*} 1 &= 4a-2b-1\\ 1 &= 9a+3b-1 \end{align*} which yields $$ 2 = 13a+b-2. $$ This gives

$$ b = -13a .$$

Now $-13$ is a strange number for a schoolbook example. And using Geogebra I found out that $a$ and $b$ must be $1/3$. In all the other questions around it, I see full number quotients and usually either $a$ or $b$ cancel themselves out, so it is easy to find out the result by substituting.

I am unsure about how to continue because I have the feeling I am missing something here.

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    $\begingroup$ You cancelled out the $2$ and $-2$, but they have opposite signs. Also, what you get is simply a system of linear equations. The easiest way to solve this would be elimination. Multiply the top equation by $3$, the bottom by $2$, and since the $b$'s are now opposite in coefficient, you can add the two equations and obtain $a$. Plug in $a$ in either equation to get $b$, and you're done. $\endgroup$ – KM101 May 20 at 16:53
  • $\begingroup$ Try taking the difference between your first two equations instead. It works out much nicer. $\endgroup$ – Michael Seifert May 20 at 16:55
  • $\begingroup$ I've edited your post; please make sure it still has what you wanted it to have. $\endgroup$ – Clayton May 20 at 16:58
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You will get the system $$2=4a-2b$$ and $$2=9a+3b$$ simplifying gives $$1=2a-b$$ $$2=9a+3b$$ Multiplying the first equation by $3$ and adding to the second we get $$5=15a$$ so $$a=\frac{1}{3}$$

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Your mistake is $b=-13a$ which should have been $b=-13a+4$

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There are various ways to deal with your simultaneous equations - you do need to be careful about signs. You can rearrange them as:

$4a-2b=2$

$9a+3b=2$

Now to eliminate $b$ multiply the first equation by $3$ (the coefficient of $b$ from the second) and the second equation by $2$ (minus the coefficient of $b$ from the first equation). You get:

$12a-6b=6$

$18a+6b=4$

And if you add these you will get $30a=10$ or $a=\frac 13$. Substituting this in the second equation gives $3+3b=2$ or $b=-\frac 13$

This is a simple mechanical method - there are often short cuts. Intelligent choices simplify arithmetic. I chose to multiply by the coefficients of $b$ rather than eliminating $a$ because these were smaller. I chose to substitute back into the second equation because $9a$ came out an integer.

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As $P_1$ and $P_2$ share the same second coordinate, the first coordinate of the parabola's vertex is the mean of their first coordinates, namely $(-2+3)/2=1/2$, on the other hand it is $-\frac{b}{2a}$. From here $b=-a$. Now plug in the coordinates of one point in $y=ax^2-ax-1$.

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