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Question Number 165830 by HongKing last updated on 09/Feb/22

\sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} - \sqrt{a \sqrt{a \sqrt{a \dots}}} = 0

(a) 9 (b) 3 (c) 6 (d) 1 (e) 12

Answered by Rasheed.Sindhi last updated on 09/Feb/22

\sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} - \sqrt{a \sqrt{a \sqrt{a \dots}}} = 0

\sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} = \sqrt{a \sqrt{a \sqrt{a \dots}}} = b (say)

8 a + \sqrt{8 a + \sqrt{8 a + \dots}} = a \sqrt{a \sqrt{a \dots}}

8 a + b = ab

8 a = b (a - 1)

b = \frac{8 a}{a - 1} \dots \dots \dots \dots (i)

Again,

\sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} = \sqrt{a \sqrt{a \sqrt{a \dots}}}

8 a + \sqrt{8 a + \sqrt{8 a + \dots}} = a \sqrt{a \sqrt{a \dots}}

\sqrt{8 a + \sqrt{8 a + \dots}} = ab - 8 a

8 a + \sqrt{8 a + \dots} = {(ab - 8 a)}^{2}

\sqrt{8 a + \dots} = {(ab - 8 a)}^{2} - 8 a

b = {(ab - 8 a)}^{2} - 8 a

\frac{8 a}{a - 1} = {(a (\frac{8 a}{a - 1}) - 8 a)}^{2} - 8 a [∵ b = \frac{8 a}{a - 1} from (i)]

\frac{8 a}{a - 1} = {(\frac{{8 a}^{2} - {8 a}^{2} + 8 a}{a - 1})}^{2} - 8 a

\frac{8 a}{a - 1} = {(\frac{8 a}{a - 1})}^{2} - 8 a

8 a (a - 1) = {64 a}^{2} - 8 a {(a - 1)}^{2}

{64 a}^{2} = 8 a (a - 1) + 8 a {(a - 1)}^{2}

= 8 a (a - 1) (1 + a - 1)

= {8 a}^{2} (a - 1)

a - 1 = 8

a = 9

Commented by Tawa11 last updated on 10/Feb/22

Weldone sir

Commented by Rasheed.Sindhi last updated on 10/Feb/22

T han k s miss!

Answered by naka3546 last updated on 09/Feb/22

\sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} - \sqrt{a \sqrt{a \sqrt{a \dots}}} = 0

\Rightarrow \sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} = \sqrt{a \sqrt{a \sqrt{a \dots}}}

\Rightarrow \sqrt{8 a + \sqrt{8 a + \sqrt{8 a + \dots}}} = a

\Rightarrow \sqrt{8 a + a} = a

\Rightarrow \sqrt{9 a} = a

\Rightarrow 9 a = a^{2} (a > 0)

\Rightarrow a = 9

Commented by Tawa11 last updated on 10/Feb/22

Weldone sir

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