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Find all real solutions to the equation \(4{x}^2 - 40\lfloor{x}\rfloor + 51 = 0\). Here, if \(x\) is a real number, then \(\lfloor{x}\rfloor\) denotes the greatest integer that is less than or equal to \(x\).

Note by Dhrubajyoti Ghosh 1 month ago

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What have you tried?

Hint: Since \(40 \lfloor x \rfloor \) and 51 are integers, then so is \(4x^2\).

Hint 2: Bound it. Express \(x\) as \(\lfloor x \rfloor + \{ x \} \). Separate the integer part and the fractional part.

Hint 3: \(x \) can be expressed as \( \frac{\sqrt A}4 \) for some integer \(A\).

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TopNewestWhat have you tried?

Hint:Since \(40 \lfloor x \rfloor \) and 51 are integers, then so is \(4x^2\).Hint 2:Bound it. Express \(x\) as \(\lfloor x \rfloor + \{ x \} \). Separate the integer part and the fractional part.Hint 3:\(x \) can be expressed as \( \frac{\sqrt A}4 \) for some integer \(A\).Log in to reply