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WAEC Further Mathematics 2019 Paper

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Question : 13

Total: 39

Find the value of x for which 6

\sqrt{4x^2+1}

= 13x, where x > 0

$\;\frac{6}{5}$
$\;\frac{25}{24}$
$\;\frac{24}{25}$
$\;\frac{5}{6}$

Solution:

$\sqrt{4x^2+1}$ = $\;\frac{13x}{6}$

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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