What is the value of \(\sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}\)?
3
Let \(x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}\). The expression repeats inside, so:
\(x = \sqrt{6 + x}\)
Squaring both sides: \(x^2 = 6 + x \;\Rightarrow\; x^2 - x - 6 = 0\).
Factorising: \((x - 3)(x + 2) = 0 \;\Rightarrow\; x = 3\ \text{or}\ x = -2\).
Since x is a square root, x must be non-negative. Hence \(x = 3\).
Similar Questions
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Rationalize:
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Important Questions from Surds and Indices
The value of \(\frac{{{{\left( {251} \right)}^3} + {{\left( {249} \right)}^3}}}{{25.1 \times 25.1 - 624.99 + 24.9 \times 24.9}}\) is 5 × 10 k , where the value of k is :
Find the value of m in \(\left(\frac{2}{7}\right)^{-3} \times \left(\frac{2}{7}\right)^{-5}=\left (\frac{2}{7}\right)^{-3m+1}\)
If √625 = 25; then√(.00000625/25)is:
A. 0.0025
B. 0.001
C. 0.0001
D. 0.0005Find the value of:
\(\sqrt{150}-\sqrt{54}-\sqrt{24}\)
If \(\sqrt{4624}=68\) , then the value of:
\(\sqrt{46.24}+\sqrt{0.4624}+\sqrt{0.004624}\)
