Question
Solve the following quadratic equation:$\text{2x}^2-\text{x}+\frac{1}{8}=0$
✓
Answer
$\text{2x}^2-\text{x}+\frac{1}{8}=0$
$\Rightarrow 16x^2 - 8x + 1 = 0$
$\Rightarrow 16x^2 - 4x - 4x + 1 = 0$
$\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0$
$\Rightarrow (4x - 1)(4x - 1) = 0$
$\Rightarrow (4x - 1)^2 = 0$
$\Rightarrow 4x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{4}$ (repeated root)
$\Rightarrow 16x^2 - 8x + 1 = 0$
$\Rightarrow 16x^2 - 4x - 4x + 1 = 0$
$\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0$
$\Rightarrow (4x - 1)(4x - 1) = 0$
$\Rightarrow (4x - 1)^2 = 0$
$\Rightarrow 4x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{4}$ (repeated root)
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