Quadratic Eq.

**~berserk** · May 27th 2008, 04:57 PM

Write the quadratic equation such that the sum of its roots is $-5$ and the product of its roots is $6$ I know that $sum=-b/a$ and $product=c/a$

**sean.1986** · May 27th 2008, 05:03 PM

Originally Posted by ~berserk

Write the quadratic equation such that the sum of its roots is $-5$ and the product of its roots is $6$ I know that $sum=-b/a$ and $product=c/a$

(x - 2)(x - 3)

x^2 -5x + 6 = 0

**Reckoner** · May 27th 2008, 06:54 PM

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 $ax^2 + bx + c$ be $x_0\text{ and }x_1$ . We have $x_0 + x_1 = -\frac ba = -5$ and $x_0x_1 = \frac ca = 6$ . This gives us the system of linear equations:

$\left\{\begin{array}{rcr} -b/a & = & -5\\ c/a & = & 6 \end{array}\right.$

$\Rightarrow\left\{\begin{array}{rcr} a & = & b/5\\ a & = & c/6 \end{array}\right.$

$\Rightarrow\frac b5 = \frac c6\Rightarrow b = \frac{5c}6$

Now, let $c$ be some arbitrary value $t$ . Then we have as our solution:

$\left.\begin{array}{rcrcr} a & = & \frac c6 & = & \frac t6\smallskip\\ b & = & \frac{5c}6 & = & \frac{5t}6\smallskip\\ c & = & t \end{array}\right\} \Rightarrow(a,\;b,\;c) = \left(\frac t6,\;\frac{5t}6,\;t\right)$

Now just pick a (nonzero) value for $t$ ! For example, with $t = 6$ we have

$ax^2 + bx + c = x^2 + 5x + 6 = (x - 2)(x - 3)$

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