Write the quadratic equation such that the sum of its roots is $\displaystyle -5$ and the product of its roots is $\displaystyle 6$ I know that $\displaystyle sum=-b/a$ and $\displaystyle product=c/a$
Hi, berserk! The easiest way to solve this problem is probably trial and error, but if you want the general solution with a more systematic approach:
Let the roots of our quadratic $\displaystyle ax^2 + bx + c$ be $\displaystyle x_0\text{ and }x_1$. We have $\displaystyle x_0 + x_1 = -\frac ba = -5$ and $\displaystyle x_0x_1 = \frac ca = 6$. This gives us the system of linear equations:
$\displaystyle \left\{\begin{array}{rcr}
-b/a & = & -5\\
c/a & = & 6
\end{array}\right.$
$\displaystyle \Rightarrow\left\{\begin{array}{rcr}
a & = & b/5\\
a & = & c/6
\end{array}\right.$
$\displaystyle \Rightarrow\frac b5 = \frac c6\Rightarrow b = \frac{5c}6$
Now, let $\displaystyle c$ be some arbitrary value $\displaystyle t$. Then we have as our solution:
$\displaystyle \left.\begin{array}{rcrcr}
a & = & \frac c6 & = & \frac t6\smallskip\\
b & = & \frac{5c}6 & = & \frac{5t}6\smallskip\\
c & = & t
\end{array}\right\}
\Rightarrow(a,\;b,\;c) = \left(\frac t6,\;\frac{5t}6,\;t\right)$
Now just pick a (nonzero) value for $\displaystyle t$! For example, with $\displaystyle t = 6$ we have
$\displaystyle ax^2 + bx + c = x^2 + 5x + 6 = (x - 2)(x - 3)$