+26400 Let a and b be the roots of \(2x^2 - x - 2 = 0\).
Find the value of \(\dfrac{1}{a} + \dfrac{1}{b}\).
\(\text{Let $\dfrac{1}{a}$ and $\dfrac{1}{b}$ be the roots of $2\left(\dfrac{1}{x}\right)^2 - \dfrac{1}{x} - 2 = 0$ }\)
\(\begin{array}{|rcll|} \hline \mathbf{2\left(\dfrac{1}{x}\right)^2 - \dfrac{1}{x} - 2} &=& \mathbf{0} \\\\ \dfrac{2}{x^2} - \dfrac{1}{x} - 2 &=& 0 \quad | \quad \cdot x^2 \\\\ 2 - x - 2x^2 &=& 0 \\\\ \mathbf{2x^2+x-2} &=& \mathbf{0} \\\\ x &=& \dfrac{-1\pm \sqrt{1^2-4*2*(-2)} } {2*2} \\\\ x &=& \dfrac{-1\pm \sqrt{17} } {4} \\\\ x_1 = \dfrac{1}{a} &=& \dfrac{-1+ \sqrt{17} } {4} \\\\ x_2 = \dfrac{1}{b} &=& \dfrac{-1- \sqrt{17} } {4} \\\\ x_1+x_2 = \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-1+ \sqrt{17} } {4}+\dfrac{-1- \sqrt{17} } {4} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-1+ \sqrt{17} -1- \sqrt{17}} {4} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-2} {4} \\\\ \mathbf{\dfrac{1}{a} + \dfrac{1}{b}} &=& \mathbf{ -\dfrac{1} {2} } \\ \hline \end{array}\)
