$2x^2 -7x + 3 = 0$ is an example for quadratic equation. Let’s solve this quadratic equation by completing the square method.

Make some basic adjustment to transform the quadratic equation as a square of a binomial.

$\implies$ $2x^2 -7x$ $=$ $-3$

In this example, the coefficient of $x^2$ is $2$. Divide both sides of the equation by the coefficient of $x^2$ term.

$\implies$ $\dfrac{2x^2 -7x}{2}$ $=$ $-\dfrac{3}{2}$

$\implies$ $\dfrac{2x^2}{2} -\dfrac{7x}{2}$ $=$ $-\dfrac{3}{2}$

$\implies$ $\Bigg(\dfrac{2}{2}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

$\implies$ $\require{cancel} \Bigg(\dfrac{\cancel{2}}{\cancel{2}}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

$\implies$ $x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

The algebraic expression in this equation can be transformed as square of sum or difference of two terms. For example ${(a+b)}^2$ or ${(a-b)}^2$.

$\implies x^2 -x\Bigg(\dfrac{7}{2}\Bigg) = -\dfrac{3}{2}$

The second term should have $2$ as a factor but there is no $2$ as factor in the second term. Therefore, include $2$ as a factor in an acceptable manner in the second term of the algebraic expression.

$\implies$ $x^2 -\Bigg(\dfrac{2}{2}\Bigg)x\Bigg(\dfrac{7}{2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2 \times \dfrac{1}{2} \times x\Bigg(\dfrac{7}{2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7 \times 1}{2 \times 2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

In the second term of the algebraic expression, the second factor $x$ represents the first term and third factor represents third factor, if this expression is compared with the expansion of square of difference of two terms but there is no term in this expression. So, add and subtract the square of this factor in the expression.

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -{\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -\dfrac{49}{16}$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{49}{16} -\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{49 -24}{16}$

The algebraic expression represents the expansion of square of difference of the two terms. So, it can be simplified mathematically as follows.

$\implies$ ${\Bigg(x -\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{25}{16}$

Now, find the values of $x$ for solving the quadratic equation.

$\implies \Bigg(x -\dfrac{7}{4}\Bigg) = \pm \sqrt{\dfrac{25}{16}}$

$\implies x -\dfrac{7}{4} = \pm \dfrac{5}{4}$

$\implies x = \dfrac{7}{4} \pm \dfrac{5}{4}$

$\implies x = \dfrac{7 \pm 5}{4}$

$\implies x = \dfrac{7 + 5}{4}$ and $x = \dfrac{7 -5}{4}$

$\implies x = \dfrac{12}{4}$ and $x = \dfrac{2}{4}$

$\,\,\, \therefore \,\,\,\,\,\, x = 3$ and $x = \dfrac{1}{2}$

Therefore, the roots of quadratic equation $2x^2 -7x + 3 = 0$ by using completing the square method are $3$ and $\dfrac{1}{2}$.

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