Solve the following quadratic equation by factorization, the roots are : $-4$, $-\dfrac{9}{4}$ $\dfrac{x \, - \, 3}{x \, + \, 3} \, - \, \dfrac{x \, + \, 3}{x \, - \, 3} \, = \, \dfrac{48}{7}, \, x \, \neq \, 3, \, x \, \neq \, -3$

Let the left side be simplified by taking the common denominator, giving (x - 3)^2 - (x + 3)^2 divided by (x^2 - 9), which equals 48/7. This simplifies to -12x / (x^2 - 9) = 48/7, and cross-multiplying yields x^2 + 7x - 9 = 0. The roots of this equation are not -4 and -9/4, because substituting x = -4 fails to satisfy the equation. The statement is False.

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