Find the roots of the equationx ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

Answer:Step-by-step explanation:Given quadratic equation:[tex]x^{2} + 3x - 8^{- 14} = 0[/tex]The solution of the given quadratic eqn is given by using Sri Dharacharya formula:[tex]x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}[/tex]The above solution is for the quadratic equation of the form:[tex]ax^{2} + bx + c = 0[/tex]   [tex]x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}[/tex]From the given eqna = 1b = 3c = [tex]- 8^{- 14}[/tex]Now, using the above values in the formula mentioned above:[tex]x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}[/tex][tex]x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})[/tex][tex]x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)[/tex]Now, Rationalizing the above eqn:[tex]x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}[/tex][tex]x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}[/tex]Solving the above eqn:[tex]x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}[/tex]Solving with the help of caculator:[tex]x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}[/tex]The precise value upto three decimal places comes out to be:[tex]x_{1, 1'} = 0.758\times 10^{- 14}[/tex]