.4. For which values of \( p \) will the equation \( x^{2}+x=p \) have no real roots?

To determine for which values of \( p \) the equation \( x^{2}+x=p \) has no real roots, we can rewrite the equation as \( x^{2} + x - p = 0 \). The roots of this quadratic equation can be found using the discriminant, which is given by \( D = b^2 - 4ac \). Here, \( a = 1 \), \( b = 1 \), and \( c = -p \). Thus, the discriminant becomes: \[ D = 1^2 - 4(1)(-p) = 1 + 4p. \] For the quadratic equation to have no real roots, the discriminant must be less than zero: \[ 1 + 4p < 0. \] Solving this inequality for \( p \): \[ 4p < -1 \implies p < -\frac{1}{4}. \] Therefore, the equation \( x^{2}+x=p \) will have no real roots for \( p < -\frac{1}{4} \).