1.

Solve the quadratic equation.\frac{2 x}{x-3}-\frac{1}{2 x+3}+\frac{3 x+9}{(x-3)(2 x+3)}=0, x \neq 3, \frac{-3}{2}

Answer»

2x / x - 3) + (1 / 2x + 3) + {3x + 9 / (x - 3)(2x + 3)} = 0

-----> {2x(2x + 3) + 1(x - 3) / (x - 3)(2x + 3)} +{3x + 9 / (x - 3)(2x + 3)} = 0

-----> {(4x^2+ 6x + x - 3) / (x - 3)(2x + 3)} +{3x + 9 / (x - 3)(2x + 3)} = 0

-----> {4x^2+ 7x - 3 / (x - 3)(2x + 3)} +{3x + 9 / (x - 3)(2x + 3)} = 0

-----> (4x^2+ 7x - 3 + 3x + 9) / (x - 3)(2x + 3) = 0

-----> 4x^2+ 10x + 6 = 0 *(x - 3)(2x + 3)

----->4x^2+ 10x + 6 = 0

-----> 2x^2+ 5x + 3 = 0

-----> 2x^2+ 2x + 3x + 3 = 0

----->2x(x + 1) + 3(x + 1) = 0

-----> (2x + 3)(x + 1) = 0

-----> x = - 3 / 2 or -1

-----> x =-1(ans



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