Using the method of completion of squares find one of the roots of the equatio

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Using the method of completion of squares find one of the roots of the equation $2 x^{2} - 7 x + 3 = 0$ . Also, find the equation obtained after completion of the square.

Options:

A . 6,

(x - \frac{7}{4})^{2} - \frac{25}{16} = 0

B . 3,

(x - \frac{7}{4})^{2} - \frac{25}{16} = 0

C . 3,

(x - \frac{7}{2})^{2} - \frac{25}{16} = 0

D . 13,

(x - \frac{7}{2})^{2} - \frac{25}{16} = 0

Answer: Option B
:
B

${2 x}^{2} - 7 x + 3 = 0$
Dividing by the coefficient of $x^{2}$ , we get
$x^{2} - \frac{7}{2} x + \frac{3}{2} = 0; a = 1, b = \frac{7}{2}, c = \frac{3}{2}$

Adding and subtracting the square of $\frac{b}{2} = \frac{7}{4}$ , (half of coefficient of x)

we get,
$[x^{2} - 2 (\frac{7}{4}) x + (\frac{7}{4})^{2}] - (\frac{7}{4})^{2} + \frac{3}{2} = 0$

The equation after completing the square is :
$(x - \frac{7}{4})^{2} - \frac{25}{16} = 0$

Taking square root, $(x - \frac{7}{4}) = (\pm \frac{5}{4})$
Taking positive sign $\frac{5}{4}, x = 3$
Taking negative sign $\frac{- 5}{4}, x = \frac{1}{2}$

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