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प्रश्न
`5x^2-4x+1=0`
उत्तर
The given equation is
`5x(x-2)+6=0`
⇒`5x^2-10x+6=0`
This is of the form `ax^2+bx+c=0` where `a=5, b=-10 and c=6`
∴ Discriminant` D=b^2-4ac=(-10)^2-4xx5xx6=100-120=-20<0`
Hence, the given equation has no real roots.
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