Find the sum and product of roots
A quadratic equation (\(x^2+bx+c=0\)) whose solutions are \(s_1\) and \(s_2\) can be written in the form: \(x^2-(s_1+s_2)x+(s_1)(s_2)=0\) where \(-b\) = the sum of the roots and \(c\) = the product of the roots.
When the equation has a leading coefficient other than \(1\), \(ax^2+bx+c=0\), the sum of the roots is \(-\large\frac{b}{a}\) and the product of the roots is \(\large\frac{c}{a}\).
Example 1: \(x^2-3x-4=0\) has a solution set of \({r, t}\). Find \(\large\frac{1}{r}+\frac{1}{t}\).
First identify the sum and product of the roots from the equation.
\(3\) is the sum of the roots and \(-4\) is the product of the roots.
Second, rewrite \(\large\frac{1}{r}+\frac{1}{t}\) as an expression of the sum and product of the roots.
\(\begin{align}&\frac{t}{t}{\cdot}\frac{1}{r}+\frac{1}{t}{\cdot}\frac{r}{r}\ & \ \ \ &\text{1) Get a common denominator.}\\
&\frac{t}{tr}+\frac{r}{tr}\ & \ \ \ &\text{2) Simplify.}\\
&\frac{t+r}{tr}\ & \ \ \ &\text{3) Simplify. What do you notice about the numerator and denominator?}\end{align}\)
Third, substitute the sum of the roots in for \(t+r\) and the product for \(tr\).
\(\large\frac{t+r}{tr}=\frac{3}{-4}\)
Therefore \(\large\frac{1}{r}+\frac{1}{t}=-\frac{3}{4}\)
Guided Learning
A quadratic equation (\(x^2+bx+c=0\)) whose solutions are \(s_1\) and \(s_2\) can be written in the form: \(x^2-(s_1+s_2)x+(s_1)(s_2)=0\) where \(-b\) = the sum of the roots and \(c\) = the product of the roots.
When the equation has a leading coefficient other than \(1\), \(ax^2+bx+c=0\), the sum of the roots is \(-\large\frac{b}{a}\) and the product of the roots is \(\large\frac{c}{a}\).
Example 1: \(x^2-3x-4=0\) has a solution set of \({r, t}\). Find \(\large\frac{1}{r}+\frac{1}{t}\).
First identify the sum and product of the roots from the equation.
\(3\) is the sum of the roots and \(-4\) is the product of the roots.
Second, rewrite \(\large\frac{1}{r}+\frac{1}{t}\) as an expression of the sum and product of the roots.
\(\begin{align}&\frac{t}{t}{\cdot}\frac{1}{r}+\frac{1}{t}{\cdot}\frac{r}{r}\ & \ \ \ &\text{1) Get a common denominator.}\\
&\frac{t}{tr}+\frac{r}{tr}\ & \ \ \ &\text{2) Simplify.}\\
&\frac{t+r}{tr}\ & \ \ \ &\text{3) Simplify. What do you notice about the numerator and denominator?}\end{align}\)
Third, substitute the sum of the roots in for \(t+r\) and the product for \(tr\).
\(\large\frac{t+r}{tr}=\frac{3}{-4}\)
Therefore \(\large\frac{1}{r}+\frac{1}{t}=-\frac{3}{4}\)
Guided Learning