Question

If ( x+1/x)= 4 find the value of

$x {}^{2} + \frac{1}{x{ {}^{2} } }$
$x {}^{4} + \frac{1} x {}^{2}$

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Answer 1

Step-by-step explanation:

$\huge\bf\orange{Given:-}$

$x + \frac{1}{x} = 4$

$\huge\bf\orange{To find out:-}$

$value \: of \: \: \: x {}^{2} + \frac{1}{x {}^{2} } \\ value \: of \: \: \: x {}^{4} + \frac{1}{x {}^{4} }$

$\huge\bf\orange{Solution:-}$

$x {}^{2} + \frac{1}{x {}^{2} } = (x + \frac{1}{x} ) {}^{2} - 2 \times x \times \frac{1}{x} \\ = 4 {}^{2} - 2 \\ = 16 - 2 \\ = 14$

$x {}^{4} + \frac{1}{x {}^{4} } = (x {}^{2} + \frac{1}{x {}^{2} }) {}^{2} - 2 \times x {}^{2} \times \frac{1}{x {}^{2} } \\ = 1</strong><strong>4</strong><strong> {}^{2} - 2 \\ = </strong><strong>1</strong><strong>9</strong><strong>6</strong><strong> - 2 \\ = </strong><strong>1</strong><strong>9</strong><strong>4$

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Answer 2

Answer:

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