Question

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

• 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 1

The correct option is B

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

${2x}^2-7x+3=0$
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 $\frac{-5}{4},x=\frac{1}{2}$