Solve the following equation using the quadra | vedclass

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(C) Given equation: $\frac{1}{x} - \frac{1}{x-2} = 3$. Taking $LCM$: $\frac{(x-2) - x}{x(x-2)} = 3$. $\frac{-2}{x^2 - 2x} = 3$. $-2 = 3x^2 - 6x$. $3x^

Vedclass · Accepted Answer

$\frac{3+\sqrt{3}}{3}, \frac{3-\sqrt{3}}{3}$

Vedclass · Answer

(C) Given equation: $\frac{1}{x} - \frac{1}{x-2} = 3$. Taking $LCM$: $\frac{(x-2) - x}{x(x-2)} = 3$. $\frac{-2}{x^2 - 2x} = 3$. $-2 = 3x^2 - 6x$. $3x^2 - 6x + 2 = 0$. Comparing with $ax^2 + bx + c = 0$,we get $a = 3, b = -6, c = 2$. Discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$. Since $D > 0$,real roots exist. Using quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$: $x = \frac{-(-6) \pm \sqrt{12}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\frac{1}{x} - \frac{1}{x-2} = 3, x \neq 0, 2$.

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