a) \(3{x^2} - 2x - 5 = 0\)
Hệ số \(a = 3, b = -2, c = -5\)
\(\eqalign{
& \Delta ' = {\left( { - 1} \right)^2} - 3.\left( { - 5} \right) = 16 > 0 \cr
& \sqrt {\Delta '} = \sqrt {16} = 4 \cr
& {x_1} = {{1 + 4} \over 3} = {5 \over 3} \cr
& {x_2} = {{1 - 4} \over 3} = - 1 \cr
& {x_1} + {x_2} = {5 \over 3} + \left( { - 1} \right) = {2 \over 3} \cr
& {x_1}{x_2} = {5 \over 3}.\left( { - 1} \right) = {{ - 5} \over 3} \cr} \)
b) \(5{x^2} + 2x - 16 = 0\)
Hệ số \(a = 5, b = 2, c = -16\)
\(\eqalign{
& \Delta ' = {1^2} - 5.\left( { - 16} \right) = 81 > 0 \cr
& \sqrt {\Delta '} = \sqrt {81} = 9 \cr
& {x_1} = {{ - 1 + 9} \over 5} = {8 \over 5} \cr
& {x_2} = {{ - 1 - 9} \over 5} = - 2 \cr
& {x_1} + {x_2} = {8 \over 5} + \left( { - 2} \right) = {{ - 2} \over 5} \cr
& {x_1}{x_2} = {8 \over 5}.\left( { - 2} \right) = {{ - 16} \over 5} \cr} \)
c) \(\displaystyle {1 \over 3}{x^2} + 2x - {{16} \over 3} = 0\)
\(\Leftrightarrow {x^2} + 6x - 16 = 0\)
Hệ số \(a = 1, b = 6, c = -16\)
\(\eqalign{
& \Delta ' = {3^2}- 1.\left( { - 16} \right) = 25 > 0 \cr
& \sqrt {\Delta '} = \sqrt {25} = 5 \cr
& {x_1} = {{ - 3 + 5} \over 1} = 2 \cr
& {x_2} = {{ - 3 - 5} \over 1} = - 8 \cr
& {x_1} + {x_2} = 2 + \left( { - 8} \right) = - 6 \cr
& {x_1}{x_2} = 2.\left( { - 8} \right) = - 16 \cr} \)
d) \(\displaystyle {1 \over 2}{x^2} - 3x + 2 = 0 \)
\(\Leftrightarrow {x^2} - 6x + 4 = 0\)
Hệ số \(a = 1, b = -6, c = 4\)
\(\eqalign{
& \Delta ' = {\left( { - 3} \right)^2} - 1.4 = 9 - 4 = 5 > 0 \cr
& \sqrt {\Delta '} = \sqrt 5 \cr
& {x_1} = {{3 - \sqrt 5 } \over 1} = 3 - \sqrt 5 \cr
& {x_2} = {{3 + \sqrt 5 } \over 1} = 3 + \sqrt 5 \cr
& {x_1} + {x_2} = 3 - \sqrt 5 + 3 + \sqrt 5 = 6 \cr} \)
\(\,\,{x_1}{x_2} = \left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right) \)\(\,= 9 - 5 = 4\).
