WAEC - Further Mathematics (2019 - No. 13)
Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x > 0
\(\frac{6}{5}\)
\(\frac{25}{24}\)
\(\frac{24}{25}\)
\(\frac{5}{6}\)
Giải thích
\(\sqrt{4x^2 + 1}\) = \(\frac{13x}{6}\)
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 = \(\pm\frac{6}{5}\)