Question

If A-1/A = 8 and A is non zero; find: A+ 1/A

Answer 1

Answer:

8.24

Explanation:

$we \: know \: that \: \\ \\ {(a - b)}^{2} = {(a + b)}^{2} - 4ab \\ \\( A - \frac{1}{A} ) {}^{2} =( A + \frac{1}{A} ) {}^{2} - 4 \times A \times \frac{1}{A} \\ \\ {8}^{2} = (A + \frac{1}{A} ) {}^{2} - 4 \times \frac{A}{A} \\\\ 64 = (A + \frac{1}{A} ) {}^{2} - 4 \\ \\64 + 4 = (A + \frac{1}{A} ) {}^{2} \\\\ 68 = (A + \frac{1}{A} ) {}^{2} \\\\ A + \frac{1}{A} = \sqrt{68} \\ \\A + \frac{1}{A} = 8.24$.

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \bold{A - \frac{1}{A} = 8}\\ \\ \underline{ \mathfrak{ \: \:To \: \: find \: :- \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \:\bold{A + \frac{1}{A} = ?}$

$\underline{ \mathfrak{ \: Now, \: }}\\ \\ \: \: \: \: \: \: \: \: \: \: \bold{ \red{A - \frac{1}{A} = 8}} \\ \\ \\ \: \: \: \: \underline{ \text{ \: Squaring both sides, we get, \: }}$

$\tt{ \: \: \: \: \: \: \: \: (A - \frac{1}{A} ) {}^{2} = (8) {}^{2} }\\ \\ \tt{ \implies \: A {}^{2} - 2 \times A \times \frac{1}{A} + (\frac{1}{A} ) {}^{2} = 64} \\ \\ \tt{ \implies \: A {}^{2} + \frac{1}{A {}^{2} } - 2 = 64 }\\ \\ \tt{\implies \: A {}^{2} - \frac{1}{A {}^{2} } = 64 + 2} \\ \\ \tt{\implies \: \red{A {}^{2} - \frac{1}{A {}^{2} } = 66}}$

$\underline{ \text{ \: Adding \red{2} in both sides, we get, \: }} \\ \\ \: \: \: \: \: \: \: \: \sf{ A {}^{2} - \frac{1}{A {}^{2} } + 2= 66 + 2 }\\ \\ \sf{\implies \: A {}^{2} - ( \frac{1}{A} ) {}^{2} + 2 \times A \times \frac{1}{A} = 68} \\ \\ \sf{\implies \: (A + \frac{1}{A} ) {}^{2} = 68 }\\ \\ \sf{\implies \: A + \frac{1}{A} = \sqrt{68}} \\ \\ \: \: \:\sf{ \red{\therefore \: \: A + \frac{1}{A} = \pm \: \sqrt{68}}}$