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:

Question

Accepted Answer

The problem involves three quadratic equations with a common negative root. Here's the step-by-step solution:

**Given Quadratic Equations:**

- $$ x^2 + mx + 9 = 0 $$
   - $$ x^2 + nx + 17 = 0 $$
   - $$ x^2 + (m+n)x + 35 = 0 $$

Let the common negative root be $$ x = p $$.

**Substitute $$ x = p $$ into the Equations:**

- From the first equation:
     $$
     p^2 + mp + 9 = 0
     $$
   - From the second equation:
     $$
     p^2 + np + 17 = 0
     $$
   - From the third equation:
     $$
     p^2 + (m+n)p + 35 = 0
     $$

**Equate Terms:**

- Subtract the first equation from the third:
     $$
     (m+n)p + 35 - mp - 9 = 0
     $$
     Simplify:
     $$
     np + 35 = 9
     $$
     Solve for $$ np $$:
     $$
     np = -26
     $$

- Subtract the second equation from the third:
     $$
     (m+n)p + 35 - np - 17 = 0
     $$
     Simplify:
     $$
     mp + 35 = 17
     $$
     Solve for $$ mp $$:
     $$
     mp = -18
     $$

**Solve for $$ m $$ and $$ n $$:**

- From $$ mp = -18 $$, substitute $$ p = -3 $$ (the given negative root):
     $$
     m(-3) = -18
     $$
     Solve for $$ m $$:
     $$
     m = 6
     $$

- From $$ np = -26 $$, substitute $$ p = -3 $$:
     $$
     n(-3) = -26
     $$
     Solve for $$ n $$:
     $$
     n = \frac{26}{3}
     $$

**Calculate $$ 2m + 3n $$:**

Substitute values of $$ m $$ and $$ n $$:
   $$
   2m + 3n = 2(6) + 3\left(\frac{26}{3}\right)
   $$
   Simplify:
   $$
   12 + 26 = 38
   $$

### Final Answer:
The value of $$ 2m + 3n $$ is **38**.  
Correct option: **38**

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