$\sqrt{7 + \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}}$
To solve the expression \(\sqrt{7 + \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}}\), we need to recognize it as an infinite nested radical. Let's assign the entire expression to a variable \(x\). Thus, we have:
\(x = \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}\)
This implies:
\(x = \sqrt{7 + x}\)
To remove the square root, we square both sides of the equation:
\(x^2 = 7 + x\)
Rearranging the terms gives us a quadratic equation:
\(x^2 - x - 7 = 0\)
We can solve this quadratic equation using the quadratic formula, where \(a = 1\), \(b = -1\), and \(c = -7\). The quadratic formula is:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substitute the values into the formula:
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-7)}}{2 \times 1}\)
Calculating further gives:
\(x = \frac{1 \pm \sqrt{1 + 28}}{2}\)
Simplifying under the square root:
\(x = \frac{1 \pm \sqrt{29}}{2}\)
The solutions to this equation are:
Since \(x\) represents a positive real number due to the context of a square root, we choose the solution:
\(x = \frac{1 + \sqrt{29}}{2}\)
Therefore, the correct answer is \(\frac{1 \pm \sqrt{29}}{2}\).
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Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is