Question

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

Answer 1

Step 1: Solve using the method of complete square

The given equation is,

$\begin{array}{l}2x^2–7x+3=0\end{array}$

$\begin{array}{l}\Rightarrow x^2–\frac{7}{2}x+\frac{3}{2}=0\end{array}$

$\Rightarrow$ $\begin{array}{l}{x}^2–\frac{7}{2}x=-\frac{3}{2}\end{array}$

Now by adding ${\left(\frac{7}{4}\right)}^2$ on both sides

$\begin{array}{l}\Rightarrow x^2–\frac{7}{2}x+{\frac{7}{4}}^2=\frac{-3}{2}+(\frac{7}{4}{)}^2\end{array}$

$\Rightarrow$ $\begin{array}{l}(x–\frac{7}{4}{)}^2=\frac{-3}{2}+\frac{49}{16}\end{array}$

$\Rightarrow$ $\begin{array}{l}(x–\frac{7}{4}{)}^2=\frac{25}{16}\end{array}$

$\Rightarrow$ $\begin{array}{l}(x–\frac{7}{4}{)}^2=(\frac{5}{4}{)}^2\end{array}$

Step 2: Determine the Roots

Now by taking square root on both sides

$\begin{array}{l}\Rightarrow (x–\frac{7}{4})=\pm \frac{5}{4}\end{array}$

$\begin{array}{l}\Rightarrow x–\frac{7}{4}=\frac{5}{4},x–\frac{7}{4}=\frac{-5}{4}\end{array}$

$\begin{array}{l}\Rightarrow x=\frac{5}{4}+\frac{7}{4},x=\frac{-5}{4}+\frac{7}{4}\end{array}$

$\begin{array}{l}\Rightarrow x=3,x=\frac{1}{2}\end{array}$

Hence, the roots of the given equation are $3$ and $\frac{1}{2}$.