+776 Using the quadratic formula, we have the two solutions \(\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}\) and \(\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}\). The reciprocals would be \(\dfrac{1}{\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}}\) and \(\dfrac{1}{\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}}\). Adding them together, we have \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).
- PM
+776 But yes, a good solution would include the use of Vieta's formula.
In the quadratic polynomial \(P(x) = ax^2 + bx + c\), the roots \(x_1\) and \(x_2\) satisfy \(x_1 + x_2 = -\dfrac{b}{a}\) and \(x_1 x_2 = \dfrac{c}{a}\).
+776 Note that in this equation, \(x_1 + x_2 = 13\) and \(x_1x_2 = 4\).
+776 So the answer is \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).
- PM
