If the equations x² + mx + 9 = 0, x2 + nx + 17 = 0 and x² + (m + n)x + 35 = 0 have a common negative root, then the value of (2m + 3n) is

The correct option is 38.

Let the common negative root of the three given quadratic equations be denoted by $\alpha$, where $\alpha <0$.

Since $\alpha$ is a root of all three equations, it must satisfy each of them. We can write the equations by substituting $x=\alpha$:

Equation (1):
${\alpha }^2+m\alpha +9=0$

Equation (2):
${\alpha }^2+n\alpha +17=0$

Equation (3):
${\alpha }^2+(m+n)\alpha +35=0$

Now, we can add Equation (1) and Equation (2) together:
$({\alpha }^2+m\alpha +9)+({\alpha }^2+n\alpha +17)=0+0$
Simplifying this sum, we get:
$2{\alpha }^2+(m+n)\alpha +26=0$

Let us call this result Equation (4). We can now subtract Equation (3) from Equation (4) to eliminate the term containing $(m+n)\alpha$:
$[2{\alpha }^2+(m+n)\alpha +26]-[{\alpha }^2+(m+n)\alpha +35]=0$
Simplifying the subtraction:
${\alpha }^2-9=0$
This gives:
${\alpha }^2=9$
Taking the square root, we have $\alpha =\pm 3$. Since we are given that the common root is negative, we select:
$\alpha =-3$

Now, substitute $\alpha =-3$ back into Equation (1) to solve for $m$:
${(-3)}^2+m(-3)+9=0$
$9-3m+9=0$
$18=3m\Rightarrow m=6$

Next, substitute $\alpha =-3$ into Equation (2) to solve for $n$:
${(-3)}^2+n(-3)+17=0$
$9-3n+17=0$
$26=3n\Rightarrow n=\frac{26}{3}$

We can verify these values in Equation (3):
${(-3)}^2+(6+\frac{26}{3})(-3)+35=9-18-26+35=0$
This confirms our calculated values of $m$ and $n$ are correct.

Finally, we calculate the required value of $(2m+3n)$:
$2m+3n=2\left(6\right)+3\left(\frac{26}{3}\right)$
$2m+3n=12+26=38$