Algebra

according to my calculations,

\(x = {-11 \pm \sqrt{11^2-4(5)(14)} \over 2(5)}\)

=\(x = {-11 \pm \sqrt{159}i \over 10}\)

Those are the two roots. 1/a and 1/b would be

\(1/a = {1/{-11 + \sqrt{159} i \over10}}\)

\(1/a = -{11+\sqrt{159} i \over 28}\)

\(1/b = {1/{-11- \sqrt{159} i \over10}}\)

\(1/b = -{11-\sqrt{159} i \over 28}\)

\(1/a+1/b= -{11+\sqrt{159} i \over 28} -{11-\sqrt{159} i \over 28} = -{22 \over 28} = -{11 \over 14}\)

Let a and b be the solutions to 5x^2 + 11x + 14 = 0.
Find 1/a + 1/b

\(\begin{array}{|rcll|} \hline \mathbf{5x^2 + 11x + 14} &=& \mathbf{0} \\ \boxed{x\rightarrow \frac1x} \\ 5\left(\dfrac{1}{x}\right)^2 + 11\left(\dfrac{1}{x}\right) + 14 &=& 0 \\\\ \dfrac{5}{x^2} + \dfrac{11}{x} + 14 &=& 0 \quad | \quad \times x^2 \\\\ 5 + 11x + 14x^2 &=& 0 \quad | \quad :14 \\\\ \dfrac{5}{14} + \dfrac{11}{14}x + x^2 &=& 0 \\\\ x^2 + \dfrac{11}{14}x + \dfrac{5}{14} &=& 0 \\\\ \boxed{\text{By Vieta:}} \\\\ x^2 + \underbrace{\dfrac{11}{14}}_{=-\left(\dfrac1a+\dfrac1b \right)}x + \underbrace{\dfrac{5}{14}}_{=\dfrac1a\times\dfrac1b} &=& 0 \\\\ \hline -\left(\dfrac1a+\dfrac1b \right) &=& \dfrac{11}{14} \\\\ \mathbf{\dfrac1a+\dfrac1b} &=& \mathbf{-\dfrac{11}{14}} \\ \hline \end{array}\)

Another way to do it is by using the root-coefficient strategy

\({1\over a} + {1\over b} = {{\alpha+\beta} \over \alpha\beta}\)

\(\alpha + \beta (-{b\over a}) = -{11\over 5}\)

\(\alpha\beta ({c\over a})={14\over 5}\)

\({-{11\over 5}\over{14\over 5}} = -{11\over 14}\)