Given $\frac{m}{n} = \frac{x^{2} + 2 y}{x^{2} - 2 y}$ , make 'x' as the subject of the formula. Then find the value of x, if $y = 2, m = 5 a n d n = 4$ .

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Solution

The correct option is A 6
Given, $\frac{m}{n} = \frac{x^{2} + 2 y}{x^{2} - 2 y}$
$m (x^{2} - 2 y) = n (x^{2} + 2 y)$
$m x^{2} - 2 m y = n x^{2} + 2 n y$
$m x^{2} - n x^{2} = 2 n y + 2 m y$
$x^{2} (m - n) = 2 y (m + n)$
$x^{2} = \frac{2 y (m + n)}{m - n}$
$x = \sqrt{\frac{2 y (m + n)}{m - n}}$
Also given, $y = 2, m = 5, n = 4$
$∴ x = \sqrt{\frac{2 \times 2 (5 + 4)}{5 - 4}} = \sqrt{36} = 6$

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Similar questions

\frac{m}{n} = \frac{x^{2} + 2 y}{x^{2} - 2 y}

x : (m^{2} - 2 n) = Y : (m^{2} + 2 n)

3 m = 5 x = y = 15

\frac{x + y - 8}{2} = \frac{x + 2 y - 14}{3} = \frac{3 x - y}{4}

x^{2} + y^{2}

\frac{10 - 3 x}{5 + 2 x} = \frac{m}{n}

x

x

3 m - 4 n = 2

n = 2.5

\frac{10 - 3 x}{5 + 2 x} = \frac{m}{n}

3 m - 411 = 2

n = 2.5

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