Finding the other $x$-intecept of a quadratic equation $ax^2+bx+c=0$ when $a$ is unknown and one $x$-intercept is known

Question

One of the $x$-intercepts of the function $f(x)=ax^2-3x+1$ is at $x=-1$. Determine $a$ and the other $x$-intercept.

I happen to know that $a=-4$ and the other $x$-intercept is at $x=\frac{1}{4}$ but I don't know how to get there. I tried substituting $x=-1$ into the quadratic formula.

$$ -1=\frac{-(-3) \pm \sqrt{(-3)^2-4a}}{2a} $$

Solving for $a$ I managed to come up with $a=-\frac {5}{2}$ by some convoluted process but obviously that doesn't work.

How do I properly solve for $a$, taking into account the known root of $x$?

Answer 1

One approach is given in comment by Winther, here is another,
$ax^2+bx+c=0$ has two roots $r_1$ and $r_2$ then $$r_1+r_2=\frac{-b}{a}$$ and $$r_1r_2=\frac{c}{a}$$

Answer 2

If $r$ and $s$ are roots of $f(x) = ax^2 + bx + c$ then $x-r$ and $x-s$ are factors of $f(x)$. Therefore, $$ f(x) = a(x - r)(x - s) = a \bigl( x^2 - (r + s)x + rs \bigr), $$ so $$ \left\{ \begin{align} -a(r + s) &= b \\ ars &= c \end{align} \right. $$ If the coefficients $b$ and $c$ are known, as well as one root $s$, then we have a system of two equations with two unknowns. Multiplying the first equation by $s$ and adding the two equations together yields $$ -as^2 = bs + c, $$ so the leading coefficient$\Large^\dagger$ is $$ a = - \frac{bs + c}{s^2}. $$ Then the other root is $$ r = \frac{c}{as} = - \frac{cs}{bs + c}. $$

$\Large^\dagger$ I am assuming that that $c \ne 0$ and, hence, that neither of the roots were zero. I leave that special (and easier) case to you.