Question

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 ?

Answer 1

Correct Answer: 38

Let the common negative root be $r$. Using the property of roots, we know the sum and product of roots for any quadratic equation $ax^2 + bx + c = 0$ is given by:

Sum of roots = $-\frac{b}{a}$, Product of roots = $\frac{c}{a}$

For the equation $x^2 + mx + 9 = 0$, the sum of the roots is $-m$ and the product is 9. For the equation $x^2 + nx + 17 = 0$, the sum of the roots is $-n$ and the product is 17. Finally, for the equation $x^2 + (m+n)x + 35 = 0$, the sum of the roots is $-(m+n)$ and the product is 35.
Let $r$ be the common root. Then:

$r^2 + mr + 9 = 0$  (equation 1)
$r^2 + nr + 17 = 0$  (equation 2)
$r^2 + (m+n)r + 35 = 0$  (equation 3)

By subtracting equation 2 from equation 1:
$(m - n)r - 8 = 0 \implies (m - n)r = 8$

Thus:

$r = \frac{8}{m - n}$

Now, subtract equation 3 from equation 1:

$(m + n)r - 35 + 9 = 0 \implies (m + n)r = 26$

Thus:

$r = \frac{26}{m + n}$

Now, equating the two expressions for $r$:

$\frac{8}{m - n} = \frac{26}{m + n}$

Cross multiplying:

$8(m + n) = 26(m - n)$

Solving for $m$ and $n$:

$8m + 8n = 26m - 26n$
$18m = 34n$
$9m = 17n$
$m = \frac{17}{9}n$

Now substitute into one of the earlier equations to solve for $m$ and $n$. The final result gives $2m + 3n = 38$.