Question

Solve by completing the square.
5x^2-2x+1=0
i just can't figure this out.

Answer 1

you have to bring it to theform (x - a)^2 = b^2

Answer 2

\[x ^{2}-\frac{ 2 }{5 }x+\frac{ 1 }{ 5 }=0\]
\[x ^{2}-\frac{ 2 }{5 }x+\left( \frac{ -2 }{ 5*2 } \right)^{2}-\left( \frac{ -2 }{ 5*2 } \right)^{2} +\frac{ 1 }{ 5} =0\]
\[\left( x-\frac{ 1 }{ 5 } \right)^{2}-\frac{ 1 }{ 25 }+\frac{ 1 }{ 5 }=0\]
\[\left( x-\frac{ 1 }{5 } \right)^{2}+\frac{ -1+5 }{ 25 }=0\]
\[\left( x-\frac{ 1 }{5 } \right)^{2}-\frac{- 4 }{ 25 }=0\]
\[\left( x-\frac{ 1 }{ 5 } \right)^{2}-\frac{ 4\iota ^{2} }{25 }=0\]
\[\left( x-\frac{ 1 }{5 } \right)^{2}-\left( \frac{ 2\iota }{ 5 } \right)^{2}=0\]
\[\left( x-\frac{ 1 }{ 5 }+ \frac{ 2\iota }{ 5 }\right)\left( x-\frac{ 1 }{ 5 }-\frac{ 2\iota }{ 5 } \right)=0\]

Answer 3

super factoring :)