Question

one upon x minus one upon x minus 2 equals to 3 find the roots of the following equation

Answer 1

Answer:

$x = \frac{-3±2\sqrt{3}}{6}$

Step-by-step explanation:

$Given \: equation :\\\frac{1}{x}-\frac{1}{x-2}=3$

$\implies \frac{x-2-x}{x(x-2)}=3$

$\implies \frac{-2}{x^{2}-2x}=3$

$\implies -2=3(x^{2}-2x)$

$\implies -2=3x^{2}-6x$

$\implies 0=3x^{2}-6x+2$

$\implies 3x^{2}-6x+2=0$

Compare this equation with ax²+bx+c=0, we get

a = 3 , b = -6 , c = 2

Discreminant (d)= b²-4ac

= (-6)²-4×3×2

= 36-24

= 12

Now ,

$x = \frac{-b±\sqrt{D}}{2a}$

$x = \frac{-3±\sqrt{12}}{2\times 3}$

$x = \frac{-3±2\sqrt{3}}{6}$

Therefore,

$x = \frac{-3±2\sqrt{3}}{6}$

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Answer 2

Answer:

hope it helps you

Step-by-step explanation:

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