Question

If
$\sf \: x = \dfrac{ \sqrt{a + 2b}+\sqrt{a - 2b}}{\sqrt{a + 2b}-\sqrt{a - 2b}}$
then, find the value of bx² - ax + b.

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Answer 1

Answer:

bx² - ax + b.=0

Step-by-step explanation:

Answer 2

☯ Explanation,

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$\displaystyle \longrightarrow \sf x = \dfrac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} }$

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$\displaystyle \longrightarrow \sf x = \dfrac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } \times \dfrac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} + \sqrt{a - 2b} }$

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$\displaystyle \longrightarrow \sf x = \dfrac{ (\sqrt{a + 2b} ) + \sqrt{a - 2b} ) {}^{2} }{ (\sqrt{a + 2b} ) {}^{2} - (\sqrt{a + 2b}) {}^{2} }$

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$\displaystyle \longrightarrow \sf x = \dfrac{( \sqrt{a + 2b}) {}^{2} + ( \sqrt{a - 2b} ) {}^{2} + 2 \times (\sqrt{a + 2b} )( \sqrt{a - 2b} )}{ \not{a }+ 2b - \not{a} + 2b}$

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$\displaystyle \longrightarrow \sf x = \dfrac{a + 2b - a + 2b + 2 ( \sqrt{a + 2b})( \sqrt{a - 2b} )}{4b}$

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$\displaystyle \longrightarrow \sf x = \dfrac{2a + 2( \sqrt{a + 2b})( \sqrt{a - 2b} )}{4b}$

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$\displaystyle \longrightarrow \sf x = \dfrac{a + (\sqrt{a + 2b})( \sqrt{a - 2b} ) }{2b}$

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$\displaystyle \longrightarrow \sf 2bx = a + ( \sqrt{a + 2b} )( \sqrt{a - 2b} )$

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$\displaystyle \longrightarrow \sf2bx - a = ( \sqrt{a + 2b} )( \sqrt{a - 2b} )$

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$\displaystyle \longrightarrow \sf \bigg(2bx - a \bigg) {}^{2} = \bigg( ( \sqrt{a + 2b} ) + ( \sqrt{a - 2b} ) \bigg) {}^{2}$

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$\displaystyle \longrightarrow \sf 4b {}^{2} x {}^{2} + a {}^{2} - 4bxa = (a + 2b)(a - 2b)$

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$\displaystyle \longrightarrow \sf 4b {}^{2} x {}^{2} + a {}^{2} - 4bxa = a {}^{2} - 4b {}^{2}$

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$\displaystyle \longrightarrow \sf 4b {}^{2} x {}^{2} + a {}^{2} - 4bxa - a {}^{2} + 4b {}^{2} = 0$

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$\displaystyle \longrightarrow \sf 4b {}^{2} x {}^{2} - 4bxa + 4b {}^{2} = 0$

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$\displaystyle \longrightarrow \sf 4b(bx {}^{2} - ax + b) = 0$

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$\displaystyle \longrightarrow \underline{ \boxed{ \sf bx {}^{2} - ax + b = 0}} \: \bigstar$