Given that \(\frac{6x+m}{2x^{2}+7x-15} \equiv \frac{4}{x+5} - \frac{2}{2x-3}\), find the value of m.
Answer Details
To solve for m, we need to first create a common denominator on the right-hand side of the equation:
\[\frac{4}{x+5} - \frac{2}{2x-3} = \frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{5}{2x-3} - \frac{3}{x+5}\]
Now we have:
\[\frac{6x+m}{2x^{2}+7x-15} = \frac{5}{2x-3} - \frac{3}{x+5}\]
Multiplying both sides by the common denominator of the right-hand side, we get:
\[(6x+m)(x+5) = 5(2x^2+7x-15) - 3(2x-3)(2x+5)\]
Expanding and simplifying, we get:
\[6x^2 + 41x + 5m + 15 = 4x^2 + 7x - 30\]
\[2x^2 + 34x + 45 + m = 0\]
Now, to find the value of m, we substitute the given roots of the equation (which are -5/2 and 9/2) into the equation and solve for m:
\[m = -2x^2 - 34x - 45\]
When x = -5/2,
\[m = -2\left(\frac{-5}{2}\right)^2 - 34\left(\frac{-5}{2}\right) - 45 = -22\]
Therefore, the value of m is -22. Option (d) is correct.