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Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$2x^2 - 7x + 3 = 0$
Given:
Given quadratic equation is $2x^2 - 7x + 3 = 0$.
To do:
We have to find the roots of the given quadratic equation.
Solution:
$2x^2 - 7x + 3 = 0$
$2(x^2 - \frac{7}{2} x +\frac{3}{2}) = 0$
$x^2 - \frac{7}{2} x +\frac{3}{2} = 0$
$x^2 - 2\times \frac{1}{2}\times \frac{7}{2} x = -\frac{3}{2}$
$x^2 - 2\times \frac{7}{4} x = -\frac{3}{2}$
Adding $(\frac{7}{4})^2$ on both sides completes the square. Therefore,
$x^2 - 2\times (\frac{7}{4}) x + (\frac{7}{4})^2 = -\frac{3}{2}+(\frac{7}{4})^2$
$(x-\frac{7}{4})^2=-\frac{3}{2}+\frac{49}{16}$ (Since $(a-b)^2=a^2-2ab+b^2$)
$(x-\frac{7}{4})^2=\frac{49-3\times8}{16}$
$(x-\frac{7}{4})^2=\frac{49-24}{16}$
$(x-\frac{7}{4})^2=\frac{25}{16}$
$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}$ (Taking square root on both sides)
$x-\frac{7}{4}=\pm \frac{5}{4}$
$x=\frac{7}{4}+\frac{5}{4}$ or $x=\frac{7}{4}-\frac{5}{4}$
$x=\frac{7+5}{4}$ or $x=\frac{7-5}{4}$
$x=\frac{12}{4}$ or $x=\frac{2}{4}$
$x=3$ or $x=\frac{1}{2}$
The values of $x$ are $3$ and $\frac{1}{2}$.