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

When we say equations have a "common root," it means there's a value that satisfies all the equations simultaneously.

Let us call this common negative root $\alpha$ (alpha). Since $\alpha$ satisfies all three equations:

$\alpha^2 + m\alpha + 9 = 0$ ... (1)

$\alpha^2 + n\alpha + 17 = 0$ ... (2)

$\alpha^2 + (m+n)\alpha + 35 = 0$ ... (3)

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Subtracting equation (1) from equation (2):

$(\alpha^2 + n\alpha + 17) - (\alpha^2 + m\alpha + 9) = 0$

The $\alpha^2$ terms cancel out:

$n\alpha - m\alpha + 17 - 9 = 0$

$(n - m)\alpha + 8 = 0$

Therefore: $\alpha = \dfrac{-8}{n - m}$ ... (4)

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Let us add equations (1) and (2) to create a useful relationship:

$(\alpha^2 + m\alpha + 9) + (\alpha^2 + n\alpha + 17) = 0$

$2\alpha^2 + (m + n)\alpha + 26 = 0$ ... (5)

Now, let us subtract equation (3) from equation (5):

$[2\alpha^2 + (m + n)\alpha + 26] - [\alpha^2 + (m + n)\alpha + 35] = 0$

Notice how the $(m + n)\alpha$ terms cancel:

$2\alpha^2 - \alpha^2 + 26 - 35 = 0$

$\alpha^2 - 9 = 0$

$\alpha^2 = 9$

$\alpha = \pm 3$

Since we're told the root is negative: $\alpha = -3$

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Substitute back into equation (4):

$-3 = \dfrac{-8}{n - m}$

$-3(n - m) = -8$

$3(n - m) = 8$

$n - m = \dfrac{8}{3}$ ... (6)

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Next, we substitute $\alpha = -3$ into equation (1):

$(-3)^2 + m(-3) + 9 = 0$

$9 - 3m + 9 = 0$

$18 = 3m$

$m = 6$

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Using equation (6) to find n:

$n - 6 = \dfrac{8}{3}$

$n = 6 + \dfrac{8}{3} = \dfrac{18 + 8}{3} = \dfrac{26}{3}$

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The question asks for $(2m + 3n)$:

$2m + 3n = 2(6) + 3\left(\dfrac{26}{3}\right) = 12 + 26 = 38$

Therefore, the value of $(2m + 3n) = 38$