Eliminate x from the equations $a + c = (b / x) - d x, a - c = (d / x) - b x$

{(b^{2} - d^{2})}^{2} = c^{2} {(b + d)}^{2} - a^{2} {(b - d)}^{2}

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{(b^{2} + d^{2})}^{2} = c^{2} {(b + d)}^{2} - a^{2} {(b - d)}^{2}

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{(b^{2} - d^{2})}^{2} = c^{2} {(b + d)}^{2} + a^{2} {(b - d)}^{2}

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{(b^{2} - d^{2})}^{2} = c^{2} {(b - d)}^{2} - a^{2} {(b - d)}^{2}

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Solution

The correct option is A ${(b^{2} - d^{2})}^{2} = c^{2} {(b + d)}^{2} - a^{2} {(b - d)}^{2}$
We have $d x^{2} + (a + c) x - b = 0, b x^{2} + (a - c) x - b = 0$
$\Rightarrow \frac{x^{2}}{- d (a + c) + b (a - c)} = \frac{x}{- (b^{2} - d^{2})} = \frac{1}{d (a - c) - b (a + c)}$
$\Rightarrow \frac{x^{2}}{a (b - d) - c (b + d)} = \frac{x}{- (b^{2} - d^{2})} = \frac{1}{- a (b - d) - c (b + d)}$
$\Rightarrow \frac{x^{2}}{c (b + d) - a (b - d)} = \frac{X}{b^{2} - d^{2}}$
$∴ M^{2} = N L$
${(b^{2} - d^{2})}^{2} = c^{2} {(b + d)}^{2} - a^{2} {(b - d)}^{2}$

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