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Answer 1

Step-by-step explanation:

Given:

$\frac{ {x}^{2} }{16} + \frac{9}{ {x}^{2} } = 2( \frac{x}{4} - \frac{3}{x} ) + \frac{1}{2}$

To find: Value of x

Solution:

Formula used:

$\boxed{( {x - y})^{2} = {x}^{2} + {y}^{2} - 2xy}$

Subtract 2(x/4)(3/x) from both sides,so that complete square will form in LHS

$\frac{ {x}^{2} }{ {(4)}^{2} } + \frac{( {3)}^{2} }{ {x}^{2} } - 2. \frac{x}{4}. \frac{3}{x} = 2( \frac{x}{4} - \frac{3}{x} ) + \frac{1}{2} - 2. \frac{x}{4}. \frac{3}{x}$

$( { \frac{x}{4} }) ^{2} +( { \frac{3}{x} }) ^{2} - 2. \frac{x}{4}. \frac{3}{x} = 2( \frac{x}{4} - \frac{3}{x} ) + \frac{1}{2} - \frac{ \cancel x}{2}. \frac{3}{ \cancel x} \\ \\ ( { \frac{x}{4} - \frac{3}{x} }) ^{2} = 2( \frac{x}{4} - \frac{3}{x} ) - 1 \\ \\ ( { \frac{x}{4} - \frac{3}{x} }) ^{2} - 2( \frac{x}{4} - \frac{3}{x} ) + 1 = 0$

its a quadratic equation in x/4-3/x

Let x/4-3/x =a

${a}^{2} - 2a + 1 = 0 \\ \\ {a}^{2} - a - a + 1 = 0 \\ \\ a(a - 1) - 1(a - 1) = 0 \\ \\ (a - 1) (a - 1)= 0 \\ \\ a - 1 = 0 \\ \\ \bold{a = 1}$

Thus,

$\frac{x}{4} - \frac{3}{x} = 1 \\ \\ \frac{ {x}^{2} - 12 }{4x} = 1 \\ \\ {x}^{2} - 12 = 4x \\ \\ {x}^{2} - 4x - 12 = 0 \\ \\ {x}^{2} - 6x + 2x - 12 = 0 \\ \\ x(x - 6) + 2(x - 6) = 0 \\ \\ (x - 6)(x + 2) = 0 \\ \\ x - 6 = 0 \\ \\ x = 6 \\ \\ or \\ \\ x + 2 = 0 \\ \\ x = - 2$

Final answer:

$\bold{x = 6} \\ \bold{x = - 2 }$

Hope it helps you.

To learn more on brainly:

Find the value of 2(a²+b²) if (a+b)=3 and (a-b)=2. Please answer quick I will surely mark u brainliest.

https://brainly.in/question/47096731

Answer 2

Answer:

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Step-by-step explanation:

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