Question

solve for x: 4/x-3/(2x+3) = 3​

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{x = \frac{ - 2 + 2 \sqrt{17} }{6} \:and\:x = \frac{ - 2 - 2\sqrt{17} }{6} }}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION- }}}}}$

>>We have to find the value of x

$= ) \frac{4}{x } - \frac{3}{2x + 3} = 3 \\ = ) \frac{4(2x + 3) - 3x}{(2x + 3)x} = 3 \\ = ) \frac{8x + 12 - 3x}{2 {x}^{2} + 3x } = 3 \\ = )5x + 12 = 6 {x}^{2} + 9x \\ = )6 {x}^{2} + 9x - 5x + 12 = 0 \\ = )6 {x}^{2} + 4x + 12 = 0 \\ = )2(3 {x}^{2} + 2x + 6) = 0 \\ = )3 {x}^{2} + 2x + 6 = 0$

>>Soving this eqn gives a quadratic eqn

>>From this quadratic we find two values of x.

$\huge{\boxed{\boxed{\underline{\sf{quadratic\:formula }}}}}$

$3 {x}^{2} + 2x + 6 = 0 \\ d = {b}^{2} - 4ac \\ d = {2}^{2} - 4(3 \times 6) \\ d = 4 - 4 \times 18 \\ d = 4 - 72 \\ d = - 68 \\ x = \frac{ - b + \sqrt{d} }{2a} \\ x = \frac{ - 2 + \sqrt{ - 68} }{6} \\ x = \frac{ - 2 - 2\sqrt{17} }{6} - - - - - 1st \: zeroes \\ x = \frac{ - 2 - \sqrt{ - 68} }{6} \\ x = \frac{ - 2 + 2 \sqrt{17} }{6} - - - - - 2nd \: zeroes$

$\huge{\boxed{\boxed{\sf{x = \frac{ - 2 + 2 \sqrt{17} }{6} \:and\:x = \frac{ - 2 - 2\sqrt{17} }{6} }}}}$

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