Question

The roots of the following quadratic equation (if they exist) by the method of completing the square.
$2x^2-7x+3=0$ are $3$, $\frac{1}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$\Rightarrow$ Here, $a=1,b=-\frac{7}{2},c=\frac{3}{2}$

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

${\left(\frac{-b}{2a}\right)}^2={\left(\frac{-\left(\frac{-7}{2}\right)}{2\left(1\right)}\right)}^2=\frac{49}{16}$

Now, adding $\frac{49}{4}$, on both sides,

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

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

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

$\Rightarrow$ $x-\frac{7}{4}=\frac{5}{4}$ and $x-\frac{7}{4}=-\frac{5}{4}$

$\Rightarrow$ $x=3$ and $x=\frac{1}{2}$