Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x >
FURTHER MATHEMATICS
WAEC 2019
Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x > 0
- A. \(\frac{6}{5}\)
- B. \(\frac{25}{24}\)
- C. \(\frac{24}{25}\)
- D. \(\frac{5}{6}\)
Correct Answer: A. \(\frac{6}{5}\)
Explanation
\(\sqrt{4x^2 + 1}\) = \(\frac{13x}{6}\)
4x\(^2\) + 1 = \(\frac{169x^2}{36}\)
4 + x\(^2\) = \(\frac{169x^2}{36}\)
cross multiply
169x\(^2\) - 144x\(^2\) = 36
25x\(^2\) = 36
x\(^2\) = \(\frac{36}{25}\)
: x = \(\pm\frac{6}{5}\)
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