SOLUTION: 5x(squared)

SOLUTION: 5x(squared) - 2x + 1 = 0

Question 29685: 5x(squared) - 2x + 1 = 0
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
5x(squared) - 2x + 1 = 0
5x^2-2x+1= 0 ----(1)
x= {-(-2)+ or - sqrt[(-2)^2-4X5X1]}/2X5
[ using given ax^2+bx+c = 0, then x={-b+or-sqrt(b^2-4ac)}/2a ]
x = (1/10){2+or-sqrt[4-20]}
=(1/10){2+or-(4i)} (sqrt(-16) = sqrt[16i^2]= sqrt4^2i^2]=4i)
Therefore x = (1/10)(2+4i) or x = (1/10)(2-4i)
That is x = (1+2i)/5 and x = (1-2i)/5 (cancelling 2 in the nr and in the dr)
Verification: Putting x = (1+2i)/5 in (1)
LHS = 5x^2-2x+1
= 5[(1+2i)^2]/25 - 2(1+2i)/5 +1
=(1/5)[1+4i^2+4i-2-4i+5]
=(1/5)[1-4+4i-2-4i+5]
=(1/5)[6-6+4i-4i]
=(1/5)X(0)
=0 =RHS
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

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