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Question
Solve the following quadratic equation:
`2x^2 - x + 1/8 = 0`
Sum
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Solution
We write, `-x = -x/2 - x/2` as `2x^2 xx 1/8 = x^2/4 = (-x/2) xx (-x/2)`
∴ `2x^2 - x + 1/8 = 0`
⇒ `2x^2 - x/2 - x/2 + 1/8 = 0`
⇒ `2x(x - 1/4) - 1/2(x - 1/4) = 0`
⇒ `(x - 1/4)(2x - 1/2) = 0`
⇒ `x - 1/4 = 0` or `2x - 1/2 = 0`
⇒ `x = 1/4` or `x = 1/4`
Hence, `1/4` is the repeated root of the given equation.
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