Question

Find the roots of the following quadratic (if they exist) by the method of completing the square.
2x²-7x+3=0

Answer 1

$\text{Given equation is}$

$2x^2-7x+3=0$

$\text{To make the coefficient of x^2 , divide by 2}$

$x^2-\frac{7}{2}x+\frac{3}{2}=0$

$x^2-\frac{7}{2}x=\frac{-3}{2}$

$\text{To make the L.H.S as a perfect square,}$$\;\text{add \frac{49}{16} on both sides}$

$x^2-\frac{7}{2}x+\frac{49}{16}=\frac{49}{16}-\frac{3}{2}$

$(x-\frac{7}{4})^2=\frac{49-24}{16}$

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

$\text{Taking square root on bothsides, we get}$

$x-\frac{7}{4}=\pm\frac{5}{4}$

$x=\frac{7}{4}\pm\frac{5}{4}$

$x=\frac{7\pm5}{4}$

$x=\frac{7+5}{4},\;\frac{7-5}{4}$

$x=3,\;\frac{1}{2}$

$\therefore\textbf{The solution set is}\;\{\bf\,3,\;\frac{1}{2}\}$