// Solving 1/x² - z = 0 avoids Quo calls and is faster, especially
	// for high precisions.
	z.sqrtInverse(z)

	// re-attach halved exponent
	return z.SetMantExp(z, b/2)
}

// Compute √x (to z.prec precision) by solving
//
//	1/t² - x = 0
//
// for t (using Newton's method), and then inverting.
func (z *Float) sqrtInverse(x *Float) {
	// let
	//   f(t) = 1/t² - x
	// then
	//   g(t) = f(t)/f'(t) = -½t(1 - xt²)