Question

What are the solutions to 6x^2-5x-4=0

Answer 1

You'd have to complete the square in order to find the solutions...


`6{ x }^( 2 )-5x-4=0\\ \\ \Rightarrow \quad 6{ x }^( 2 )-5x=4\\ \\ \Rightarrow \quad \frac { 6{ x }^( 2 ) }{ 6 } -\frac { 5x }{ 6 } =\frac { 4 }{ 6 } \\ \\ \Rightarrow \quad { x }^( 2 )-\frac { 5 }{ 6 } x=\frac { 2 }{ 3 }`


`\\ \\ \Rightarrow \quad { \left( x-\frac { 5 }{ 12 } \right) }^( 2 )-{ \left( \frac { 5 }{ 12 } \right) }^( 2 )=\frac { 2 }{ 3 } \\ \\ \Rightarrow \quad { \left( x-\frac { 5 }{ 12 } \right) }^( 2 )-\frac { 25 }{ 144 } =\frac { 2 }{ 3 }`


`\\ \\ \Rightarrow \quad { \left( x-\frac { 5 }{ 12 } \right) }^( 2 )=\frac { 2 }{ 3 } +\frac { 25 }{ 144 } \\ \\ \Rightarrow \quad { \left( x-\frac { 5 }{ 12 } \right) }^( 2 )=\frac { 96 }{ 144 } +\frac { 25 }{ 144 } \\ \\ \Rightarrow \quad { \left( x-\frac { 5 }{ 12 } \right) }^( 2 )=\frac { 121 }{ 144 }`


`\\ \\ \Rightarrow \quad x-\frac { 5 }{ 12 } =\pm \sqrt { \frac { 121 }{ 144 } } \\ \\ \Rightarrow \quad x-\frac { 5 }{ 12 } =\pm \frac { 11 }{ 12 } \\ \\ \Rightarrow \quad x=\frac { 5 }{ 12 } \pm \frac { 11 }{ 12 } \\ \\ \therefore \quad x=\frac { 16 }{ 12 } =\frac { 4 }{ 3 } \\ \\ \therefore \quad x=-\frac { 6 }{ 12 } =-\frac { 1 }{ 2 }`