Common Roots Condition for Quadratic Equations

Given two quadratic equations:

1. $\mathrm{(a^2 - 7a)x^2 + x + 12 = 0}$
2. $240x^2 + 8x + (a^2 - 4) = 0$

For two quadratic equations $\mathrm{A_1x^2 + B_1x + C_1 = 0}$ and $\mathrm{A_2x^2 + B_2x + C_2 = 0}$ to have common roots, their coefficients must be proportional:

$ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} $

Determining the Value of 'a'

Applying the condition to the given equations:

$ \frac{a^2 - 7a}{240} = \frac{1}{8} = \frac{12}{a^2 - 4} $

From the first equality $\mathrm{\frac{a^2 - 7a}{240} = \frac{1}{8}}$:

$ 8(a^2 - 7a) = 240 $

$ a^2 - 7a = \frac{240}{8} $

$ a^2 - 7a = 30 $

$ a^2 - 7a - 30 = 0 $

Factorizing the equation for 'a':

$ (a - 10)(a + 3) = 0 $

Since 'a' is given as a positive integer, we have $\mathrm{a = 10}$.

Verification: Let's check if $\mathrm{a = 10}$ satisfies the second equality $\mathrm{\frac{1}{8} = \frac{12}{a^2 - 4}}$:

$ \frac{12}{10^2 - 4} = \frac{12}{100 - 4} = \frac{12}{96} = \frac{1}{8} $

The condition holds true.

Calculating the Final Expression

We need to find the value of the expression $\mathrm{\frac{(a + 2)(a + 3)}{a - 4}}$.

Substitute $\mathrm{a = 10}$ into the expression:

$ \frac{(10 + 2)(10 + 3)}{10 - 4} = \frac{(12)(13)}{6} $

$ = \frac{156}{6} $

$ = 26 $

The value of the expression is 26.