Find the roots of the following quadratic equ | vedclass

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Question

(A) 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 + \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given equation: $2x^{2} - 7x + 3 = 0$.Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} - \frac{7}{2}x + \frac{3}{2} = 0$.Shift the constant term to the right side:$x^{2} - \frac{7}{2}x = -\frac{3}{2}$.Add the square of half the coefficient of $x$ (which is $\frac{1}{2} 	imes \frac{7}{2} = \frac{7}{4}$) to both sides:$x^{2} - \frac{7}{2}x + (\frac{7}{4})^{2} = -\frac{3}{2} + (\frac{7}{4})^{2}$.$(x - \frac{7}{4})^{2} = -\frac{3}{2} + \frac{49}{16}$.$(x - \frac{7}{4})^{2} = \frac{-24 + 49}{16} = \frac{25}{16}$.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 roots are $3$ and $\frac{1}{2}$.

Find the roots of the following quadratic equation by the method of completing the square: $2x^{2} - 7x + 3 = 0$.

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