Question:medium

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 ?

Updated On: Jan 13, 2026
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Correct Answer: 38

Solution and Explanation

Let the common negative root be $r$. For any quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.

For $x^2 + mx + 9 = 0$, the sum of roots is $-m$ and the product is 9.

For $x^2 + nx + 17 = 0$, the sum of roots is $-n$ and the product is 17.

For $x^2 + (m+n)x + 35 = 0$, the sum of roots is $-(m+n)$ and the product is 35.
Since $r$ is the common root, it satisfies all three equations:

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

Subtract equation 2 from equation 1:

$(m - n)r - 8 = 0 \implies (m - n)r = 8$

Therefore:

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

Subtract equation 3 from equation 1:

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

Therefore:

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

Equating the two expressions for $r$:

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

Cross-multiply:

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

Solve for $m$ and $n$:

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

Substitute this relationship into one of the earlier equations to find $m$ and $n$. The final result is $2m + 3n = 38$.

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