Roots of Quadratic EquationsThis is a featured page

We recall that for ax^2 + bx + c = 0 whose roots are \alpha and \beta, sum of roots \alpha + \beta = -\frac{b}{a} and product of roots \alpha\beta = \frac{c}{a}.

It is a common difficulty for students to form quadratic equations whose roots are more complicated, e.g. \alpha + \frac{1}{\alpha}, \beta + \frac{1}{\beta}.

The strategy is to find the sum and product of the new roots and make use of the original values of \alpha + \beta, \alpha\beta and a very useful identity \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta.

Let's apply the above method for our example: The equation 2x^2 - 4x - 13 = 0 has roots \alpha and \beta. Form an equation whose roots are \alpha + \frac{1}{\alpha} and \beta + \frac{1}{\beta}.

Step 1: Find \alpha + \beta and \alpha\beta based on the original equation.
\alpha + \beta = -\left(\frac{-4}{2}\right) = 2
\alpha\beta = -\frac{13}{2}

Step 2: Find sum and product of new roots.
\left(\alpha + \frac{1}{\alpha}\right) + \left(\beta + \frac{1}{\beta}\right) = \left(\alpha + \beta\right) + \left(\frac{1}{\alpha} + \frac{1}{\beta} \right)
= \left(\alpha + \beta\right) + \left(\frac{\beta + \alpha}{\alpha\beta} \right)
= \frac{22}{13}

\left(\alpha + \frac{1}{\alpha} \right)\left(\beta + \frac{1}{\beta} \right) = \alpha\beta + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{1}{\alpha\beta}
= \alpha\beta + \frac{\alpha^2 + \beta^2}{\alpha\beta} + \frac{1}{\alpha\beta}
= \alpha\beta + \frac{\left(\alpha + \beta\right)^2 - 2\alpha\beta}{\alpha\beta} + \frac{1}{\alpha\beta}
= -\frac{241}{26}

Step 3: Form the new equation.
The required equation is x^2 - \frac{22}{13}x - \frac{241}{26} = 0.



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