If the roots of the following quadratic equat | vedclass

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Question

(B) Given equation: $\frac{2}{x^{2}}-\frac{1}{x}=6$.Multiply the entire equation by $x^{2}$ (where $x 
eq 0$): $2 - x = 6x^{2}$.Rearrange into standa

Vedclass · Accepted Answer

$-\frac{2}{3}, \frac{1}{2}$

Vedclass · Answer

(B) Given equation: $\frac{2}{x^{2}}-\frac{1}{x}=6$.Multiply the entire equation by $x^{2}$ (where $x 
eq 0$): $2 - x = 6x^{2}$.Rearrange into standard form: $6x^{2} + x - 2 = 0$.Divide by $6$: $x^{2} + \frac{1}{6}x - \frac{1}{3} = 0$.To complete the square,add and subtract $(\frac{1}{2} 	imes 	ext{coefficient of } x)^{2} = (\frac{1}{2} 	imes \frac{1}{6})^{2} = (\frac{1}{12})^{2} = \frac{1}{144}$.$x^{2} + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144} - \frac{1}{3} = 0$.$(x + \frac{1}{12})^{2} = \frac{1}{144} + \frac{1}{3} = \frac{1 + 48}{144} = \frac{49}{144}$.Taking the square root: $x + \frac{1}{12} = \pm \frac{7}{12}$.Case $1$: $x = \frac{7}{12} - \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$.Case $2$: $x = -\frac{7}{12} - \frac{1}{12} = -\frac{8}{12} = -\frac{2}{3}$.Thus,the roots are $-\frac{2}{3}, \frac{1}{2}$.

If the roots of the following quadratic equation exist,find them by the method of completing the square: $\frac{2}{x^{2}}-\frac{1}{x}=6$.

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