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If the equation $x^{2} - x + m^{2} = 0$ has no real roots, then $' m'$ satisfies

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$m < - \frac{1}{2}$

$m = 0$

$m < \frac{1}{2}$

$m = 2$

answer is A.

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Detailed Solution

$x^{2} - x + m^{2} = 0$ has no real roots

$Δ < 0$

$b^{2} - 4 a c < 0$

$1 - 4 m^{2} < 0$

$4 m^{2} - 1 > 0$

$(2 m + 1) (2 m - 1) > 0$

$m < \frac{- 1}{2}, m > \frac{1}{2}$

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