If $f\left(x\right)=(x^2+3x)(x^2+3x+2)$, then the sum of all real roots of the equation
$\sqrt{f\left(x\right)}+1=9701$

The correct option is -3.

Step 1: Simplify the expression for $f\left(x\right)$
We are given:

$f\left(x\right)=(x^2+3x)(x^2+3x+2)$

Let us introduce a substitution to make the expression simpler. Set $u=x^2+3x$.
Then, the function becomes:

$f\left(x\right)=u(u+2)=u^2+2u$

Step 2: Substitute $f\left(x\right)$ into the given equation
The given equation is:

$\sqrt{f\left(x\right)}+1=9701$

Subtract $1$ from both sides:

$\sqrt{f\left(x\right)}=9700$

Squaring both sides gives:

$f\left(x\right)={9700}^2$

Step 3: Solve for $u$
Replacing $f\left(x\right)$ with $u^2+2u$:

$u^2+2u={9700}^2$

Add $1$ to both sides to complete the square:

$u^2+2u+1={9700}^2+1$

${(u+1)}^2={9700}^2+1$

Taking the square root on both sides:

$u+1=\pm \sqrt{{9700}^2+1}$

Thus, we get two values for $u$:

$u=-1\pm \sqrt{{9700}^2+1}$

Step 4: Analyze the real roots of $x$
Recall that $u=x^2+3x$. Therefore, we have two quadratic equations in $x$:

1) $x^2+3x-u_1=0$, where $u_1=-1+\sqrt{{9700}^2+1}$
2) $x^2+3x-u_2=0$, where $u_2=-1-\sqrt{{9700}^2+1}$

For a quadratic equation of the form $x^2+3x-u=0$, the discriminant is:

$D=3^2-4\left(1\right)(-u)=9+4u$

Let's check the real roots condition ($D\ge <!--\; \ge \; -->0$) for both values:

- For $u_1=-1+\sqrt{{9700}^2+1}$: Since $\sqrt{{9700}^2+1}>9700$, $u_1>9699$, so $D_1=9+4u_1>0$. This equation yields two real roots.
- For $u_2=-1-\sqrt{{9700}^2+1}$: Here $u_2<-9701$, so $D_2=9+4u_2<0$. This equation gives no real roots.

Step 5: Calculate the sum of the real roots
The real roots come exclusively from the quadratic equation:

$x^2+3x-u_1=0$

By Vieta's formulas, for any quadratic equation $a{x}^2+bx+c=0$, the sum of the roots is $-\frac{b}{a}$.
Here, $a=1$ and $b=3$.

Therefore, the sum of all real roots is:

$\text{Sum of roots}=-\frac{3}{1}=-3$