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 $\displaystyle product=c/a$

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- May 27th 2008, 04:57 PM~berserkQuadratic Eq.
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 $\displaystyle product=c/a$

- May 27th 2008, 05:03 PMsean.1986
- May 27th 2008, 06:54 PMReckoner
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

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