Question

Solve 1/(x-2)+ 1/(x+3)= 1/2, x is not equal to 2 and -3


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Answer 1

Step-by-step explanation:

Given equation is 1/(x-2) + 1/(x+3)= 1/2

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

=> 2x+1 / x²+3x-2x-6 = 1/2

under cross multiplication,

2(2x+1) = x²+3x-2x-6

=> 4x+2 = x²+x-6

=> x²+x-6-4x-2 = 0

=> x²-3x-8 = 0

=> x = -b±√b²-4ac / 2a

=> x = -(-3)±√(-3)²-4(1)(-8) / 2(1)

=> x = 3±√9+32 / 2

=> x = 3±√41 / 2

.•. > x = 3+√41 / 2 or x = 3-√41 / 2

Answer 2

$\sf\huge\underline{Solution :-}$

$\sf \dfrac{1}{x - 2} + \dfrac{1}{x - 3} = \dfrac{1}{2}$

$\sf\longrightarrow \dfrac{(x + 3) + (x - 2)}{(x - 2)(x + 3)} = \dfrac{1}{2}$

$\sf\longrightarrow \dfrac{x + 3 + x - 2}{{x}^{2} + 3x - 2x - 6} = \dfrac{1}{2}$

$\sf\longrightarrow \dfrac{x + 3 + x - 2}{{x}^{2} + x - 6} = \dfrac{1}{2}$

$\sf\longrightarrow \dfrac{2x + 1}{{x}^{2} + x - 6} = \dfrac{1}{2}$

$\sf Cross \: multiplying \: the \: terms.$

$\sf\longrightarrow {x}^{2} + x - 6 = 2(2x + 1)$

$\sf\longrightarrow {x}^{2} + x - 6 = 4x + 2$

$\sf Transposing \: the \: terms \: to \: the \: left \: side.$

$\sf\longrightarrow {x}^{2} + x - 6 - 4x - 2 = 0$

$\sf\longrightarrow {x}^{2} - 3x - 8 = 0$

$\sf Since, \: we \: can't \: split \: the \: middle \: term,$

$\sf By \: the \: Quadratic \: formula,$

$\sf\star x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

[where, a = 1, b = -3 and c = -8]

$\sf\longrightarrow x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\sf\longrightarrow x = \dfrac{ -(-3) \pm \sqrt{ {-3}^{2} - 4 \times 1 \times (-8) } }{2 \times 1}$

$\sf\longrightarrow x = \dfrac{3 \pm \sqrt{ 9 - (-32)} }{2}$

$\sf\longrightarrow x = \dfrac{3 \pm \sqrt{41} }{2}$

$\sf\therefore x = \dfrac{3 + \sqrt{41} }{2} or, \dfrac{3 - \sqrt{41} }{2}$