求常数 $k$ 的取值范围，使得 $k\left(x^2 + x\right) + 2\left(x^2 + x + 1\right) = 0$ 无实数根。 由此说明为什么 $-\left(x^2 + x\right) + 2\left(x^2 + x + 1\right)$ 恒大于零。

1. Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$

$k(x^2 + x) + 2(x^2 + x + 1) = 0$
$kx^2 + kx + 2x^2 + 2x + 2 = 0$
$(k + 2)x^2 + (k + 2)x + 2 = 0$

2. Apply the discriminant condition for no real roots

For no real roots, $b^2 - 4ac < 0$.
$(k + 2)^2 - 4(k + 2)(2) < 0$

3. Simplify the inequality

$(k + 2)[(k + 2) - 8] < 0$
$(k + 2)(k - 6) < 0$

4. Solve the quadratic inequality to find the range of $k$

$-2 < k < 6$

1. Relate the given expression to the equation in part (a)

The expression $-\left(x^2 + x\right) + 2\left(x^2 + x + 1\right)$ corresponds to the left-hand side of the equation in part (a) when $k = -1$.

2. Use the condition for no real roots and the coefficient of $x^2$ to explain why it is always positive

From part (a), the equation has no real roots for $-2 < k < 6$. Since $k = -1$ is within this range, $-\left(x^2 + x\right) + 2\left(x^2 + x + 1\right) = 0$ has no real roots.
Simplifying the expression gives $x^2 + x + 2$.
Since the coefficient of $x^2$ is $1$, which is greater than $0$, the quadratic curve opens upwards ($U$-shaped). Having no real roots means it does not intersect the $x$-axis. Therefore, the expression is always positive.