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5x(squared) - 2x + 1 = 0
\n" ); document.write( "5x^2-2x+1= 0 ----(1)
\n" ); document.write( "x= {-(-2)+ or - sqrt[(-2)^2-4X5X1]}/2X5
\n" ); document.write( "[ using given ax^2+bx+c = 0, then x={-b+or-sqrt(b^2-4ac)}/2a ]
\n" ); document.write( "x = (1/10){2+or-sqrt[4-20]}
\n" ); document.write( "=(1/10){2+or-(4i)} (sqrt(-16) = sqrt[16i^2]= sqrt4^2i^2]=4i)
\n" ); document.write( "Therefore x = (1/10)(2+4i) or x = (1/10)(2-4i)
\n" ); document.write( "That is x = (1+2i)/5 and x = (1-2i)/5 (cancelling 2 in the nr and in the dr)
\n" ); document.write( "Verification: Putting x = (1+2i)/5 in (1)
\n" ); document.write( "LHS = 5x^2-2x+1
\n" ); document.write( "= 5[(1+2i)^2]/25 - 2(1+2i)/5 +1
\n" ); document.write( "=(1/5)[1+4i^2+4i-2-4i+5]
\n" ); document.write( "=(1/5)[1-4+4i-2-4i+5]
\n" ); document.write( "=(1/5)[6-6+4i-4i]
\n" ); document.write( "=(1/5)X(0)
\n" ); document.write( "=0 =RHS
\n" ); document.write( "Since this complex root holds and as complex roots occur in conjugate pairs,there is no need to try the validity of the other complex root\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "