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(D) Given equation: $2x^2 - 7x + 3 = 0$Divide the entire equation by $2$ to make the coefficient of $x^2$ equal to $1$:$x^2 - \frac{7}{2}x + \frac{3}{

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$\frac{1}{2}$ or $3$

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(D) Given equation: $2x^2 - 7x + 3 = 0$Divide the entire equation by $2$ to make the coefficient of $x^2$ equal to $1$:$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} 	imes \frac{7}{2})^2 = (\frac{7}{4})^2 = \frac{49}{16}$:$x^2 - \frac{7}{2}x + \frac{49}{16} - \frac{49}{16} + \frac{3}{2} = 0$$(x - \frac{7}{4})^2 - \frac{49}{16} + \frac{24}{16} = 0$$(x - \frac{7}{4})^2 - \frac{25}{16} = 0$$(x - \frac{7}{4})^2 = (\frac{5}{4})^2$Taking the square root on both sides:$x - \frac{7}{4} = \pm \frac{5}{4}$Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$Thus,the solutions are $x = 3$ or $x = \frac{1}{2}$.

Solve the following quadratic equation using the method of 'completing the square': $2x^2 - 7x + 3 = 0$

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