If the roots of the following quadratic equat | vedclass

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Question

(D) Given equation: $3x - \frac{2}{x} - 7 = 0$.Multiply by $x$ to get the standard form: $3x^2 - 7x - 2 = 0$.Divide by $3$: $x^2 - \frac{7}{3}x - \fra

Vedclass · Accepted Answer

$\frac{7+\sqrt{73}}{6}, \frac{7-\sqrt{73}}{6}$

Vedclass · Answer

(D) Given equation: $3x - \frac{2}{x} - 7 = 0$.Multiply by $x$ to get the standard form: $3x^2 - 7x - 2 = 0$.Divide by $3$: $x^2 - \frac{7}{3}x - \frac{2}{3} = 0$.Add and subtract $(\frac{1}{2} 	imes 	ext{coefficient of } x)^2 = (\frac{7}{6})^2 = \frac{49}{36}$:$x^2 - \frac{7}{3}x + \frac{49}{36} - \frac{49}{36} - \frac{2}{3} = 0$.$(x - \frac{7}{6})^2 = \frac{49}{36} + \frac{24}{36} = \frac{73}{36}$.Taking square root: $x - \frac{7}{6} = \pm \frac{\sqrt{73}}{6}$.$x = \frac{7 \pm \sqrt{73}}{6}$.

If the roots of the following quadratic equation exist,find them by the method of completing the square: $3x - \frac{2}{x} - 7 = 0$.

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