Question

3a+1/3a=2 root 3 find 3a-1/3a and 9a square +1/9a square

Answer 1

$3a + \frac{1}{3a} = 2\sqrt{3} \: ---(1)$

$\Big( 3a - \frac{1}{3a}\Big)^{2} \\= \Big( 3a + \frac{1}{3a}\Big)^{2} - 4 \times (3a)\times \frac{1}{3a}\\= (2\sqrt{3})^{2} - 4 \\= 12 - 4 \\= 8$

$\implies \Big( 3a - \frac{1}{3a}\Big)= \sqrt{8} \\= 2\sqrt{2} \: --(2)$

$Now, 9a^{2} + \frac{1}{9a^{2}}\\= (3a)^{2} + \frac{1}{(3a)^{2}}\\= \Big(3a + \frac{1}{3a}\Big)^{2} - 2 \times (3a) \times \frac{1}{(3a)} \\= (2\sqrt{3})^{2} - 2 \\= 12 - 2 \\= 10$

Therefore.,

$\red { Value \: of \: 9a^{2} + \frac{1}{9a^{2}}} \green { = 10 }$

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Answer 2

answer 10

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