Question

Solve the following equation using completing the square method.

$2x^2-x+\frac{1}{8}=0$

• A. $x=-\frac{1}{4}$
• B. $x=\frac{1}{4},\frac{1}{4}$
• C. $x=\frac{3}{4},\frac{3}{4}$
• D. $x=-\frac{1}{4},-\frac{1}{4}$

Answer 1

The correct option is

B

$x=\frac{1}{4},\frac{1}{4}$

$2x^2-x+\frac{1}{8}=0$

Divide throughout by 2

$\Rightarrow x^2-\frac{1}{2}x+\frac{1}{16}=0$

$\Rightarrow x^2-\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+\frac{1}{16}=0$

$\Rightarrow x^2-\frac{1}{2}x+\frac{1}{16}-\frac{1}{16}+\frac{1}{16}=0$

$\Rightarrow (x-\frac{1}{4}{)}^2-\frac{1}{16}+\frac{1}{16}=0$

$\Rightarrow (x-\frac{1}{4}{)}^2=0$

$\Rightarrow x-\frac{1}{4}=0$

$\Rightarrow x=\frac{1}{4}$

Therefore, $\frac{1}{4}$ and $\frac{1}{4}$ are the roots of the given quadratic equation.