The Square Root Of 7 Derivation

x=\sqrt { 7 } \\ \\ { x }^{ 2 }=7\\ \\ { x }^{ 2 }+2x=7+2x\\ \\ x\left( 2+x \right) =2\left( 2+x \right) +3\\ \\ \frac { x\left( 2+x \right) }{ \left( 2+x \right) } =\frac { 2\left( 2+x \right) }{ \left( 2+x \right) } +\frac { 3 }{ \left( 2+x \right) } \\ \\ x=2+\frac { 3 }{ 2+x } \\ \\ \sqrt { 7 } =2+\frac { 3 }{ 2+x } \\ \\ =2+\frac { 3 }{ 2+\left( 2+\frac { 3 }{ 2+x } \right) }

\\ \\ =2+\frac { 3 }{ 4+\frac { 3 }{ 2+x } } \\ \\ =2+\frac { 3 }{ 4+\frac { 3 }{ 2+\left( 2+\frac { 3 }{ 2+x } \right) } } \\ \\ =2+\frac { 3 }{ 4+\frac { 3 }{ 4+\frac { 3 }{ 2+x } } } \\ \\ \therefore \sqrt { 7 } =2+\frac { 3 }{ 4+\frac { 3 }{ 4+\frac { 3 }{ 4+... } } }

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