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\((9 + 4√5)^{(x^2-3)} + (9- 4√5)^{(x^2-3)} = 18\) ; Then find the number of possible integral values of x?

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Correct Answer - Option 3 : 2

 Calculation:

\((9 + 4√5)^{(x^2-3)} + (9- 4√5)^{(x^2-3)} = 18\) possible if and only if power of both the term are 1 or – 1 as (9 + 4√5)(9 + 4√5) = 1 & (9 + 4√5) + (9 + 4√5) = 18

\(⇒ (x^2-3) = 1 \)

\(⇒x^2 = 4\)

\(⇒(x^2-3) = -1 \)

\(⇒x^2 = 2\)

Therefore, x will take 4 Values 2, -2, √2, -√2 but there are only two integral value 2 and -2

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