Question

$\tt{If{\left( \sqrt{(49 + 20 \sqrt{6} )} \right)}^{ \sqrt{a \sqrt{a \sqrt{a... \infty} } } } + {\left( 5 - 2 \sqrt{6} \right)}^{ {x}^{2} + x - 3 - \sqrt{x \sqrt{x \sqrt{x... \infty } } } } =10 }$
$\tt{where } \\ \\ \tt{a = {x}^{2} - 3} \\ \\ \textsf{then \: x \:is}$
​

Answer 1

Answer:

a=I

so

x² - 3= I

x² =4

x= 2 or x =-2

Step-by-step explanation:

Refer to the photo

Also,

infinite sum of GP= a/I-r or a/r-I

(depends on value of r)

the exponents on "a" form a geometric progression with common ratio I/2.

Answer 2

x =  2   or  √2

Step-by-step explanation:

let say  b  = √a√a√a√a ............∞

Squaring both sides

=> b² = a √a√a√a√a ............∞

=> b² = ab

=> b(b - a) = 0

b = a   as b ≠0

=> √a√a√a√a ............∞ = a

Simialrly

√x√x√x√x ............∞ = x

x² + x - 3 - x  = x² - 3  

√(49 + 20√6)  = √(5 + 2√6)² = (5 + 2√6)

(5 + 2√6)ᵃ  + (5 - 2√6)ᵃ    = 10

=> (5 + 2√6)ᵃ = y

as 5 + 2√6  = 1/(5 - 2√6)

=> y  + 1/y  = 10

=> y² + 1 = 10y

=> y² - 10y + 1 =0

=> y = (10 ± √96)/2

=> y = 5 ± 2√6

(5 + 2√6)ᵃ = 5 + 2√6

=> a = 1

=> x² - 3 = 1

=> x² = 4

=> x =  2    ( x ≠ - 2  as root of -ve not defined)

(5 + 2√6)ᵃ = 5 - 2√6

=> a = -1

=>  x² - 3 = -1

=> x² = 2

=> x = √2

x =  2   or  √2

Learn more:

\mathsf{ {(2 + \sqrt{3} )}^{ {x}^{2} - 2x + 1} + {(2 - \sqrt{...

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