<s>[INST] Quadratic equation where constant term is an integral function Consider a quadratic equation ${x^2} + 2x = k + \int\limits_0^1 {\left| {t + k} \right|dt}$. Then choose the correct option(s),
(A) Roots are Real
(B) Roots are Imaginary
(C) Roots are Distinct
(D) Roots are complex number
My approach is as follow
Let $-c=k + \int\limits_0^1 {\left| {t + k} \right|dt}$
$a=1$ & $b=2$
$b^2-4ac$
$4-4(k + \int\limits_0^1 {\left| {t + k} \right|dt})$
How do I check the nature of the roots
 [/INST] The discriminant $b^2 - 4ac$ is actually
$$d = 4 + 4(k + \int_{0}^{1}\left| t + k \right| dt) \tag{1}\label{eq1A}$$
Regarding calculating $k + \int_{0}^{1}\left| t + k \right| dt$, there are $3$ cases to consider:
Case $1$: $\; k \lt -1$
Since $t + k \lt 0$ for $t$ from $0$ to $1$, this gives
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(-t - k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{1} \\
& = k + \left(-\frac{1}{2} - k\right) \\
& = -\frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
In \eqref{eq1A}, this gives $d = 2$.
Case $2$: $\; -1 \le k \lt 0$
This requires splitting the integration into $2$ parts, the initial one for $t$ from $0$ to $-k$ where $t + k \le 0 \implies \left|t + k\right| = -t -k$ and a second part for $t$ from $-k$ to $1$ where $t + k \ge 0 \implies \left|t + k\right| = t + k$, giving
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{-k}(-t - k) dt + \int_{-k}^{1}(t + k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{-k} + \left. \left(\frac{t^2}{2} + kt\right)\right|_{-k}^{1} \\
& = k - \frac{k^2}{2} + k^2 + \frac{1}{2} + k - \frac{k^2}{2} + k^2 \\
& = 2k + k^2 + \frac{1}{2} \\
& = k^2 + 2k + 1 - 1 + \frac{1}{2} \\
& = (k + 1)^2 - \frac{1}{2}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
For $k \ge -1$, the value above is an increasing function, going from $-\frac{1}{2}$ to $\frac{1}{2}$. Thus, $d$ in \eqref{eq1A} goes from $2$ to $6$.
Case $3$: $\; k \ge 0$
This obviously gives $d \gt 0$, but doing the calculations results in
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(t + k) dt \\
& = k + \left. \left(\frac{t^2}{2} + kt\right)\right|_{0}^{1} \\
& = k + \left(\frac{1}{2} + k\right) \\
& = 2k + \frac{1}{2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, $d$ in \eqref{eq1A} is $6$ at $k = 0$, and increases without bound as $k$ goes up.

This shows all $3$ cases gives $d \gt 0$, so the roots are real and distinct. Since the roots are also always complex numbers, this means options (A), (C) and (D) are true, with only (B) being false since there are no imaginary roots.
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