Solve (5+2√6)^(x^2-3) + (5-2√6)^(x^2 - 3) = 10 by using quadratic formula.

Problem : Solve (5+2√6)^(x2-3) + (5-2√6)^(x2-3) = 10 by using quadratic formula.

Doubt by Muskan 

Solution :  

(5+2√6)^(x2-3) + (5-2√6)^(x2-3) = 10 ---------(1)

Let y = (5 + 2√6)^{(x2 - 3)} 

then 1/y = 1/[(5 + 2√6)^{(x2 - 3)} ]

1/y = [1/(5+2√6)]^{(x2 - 3)} 

[On Rationalising the denominator]

1/y = (5 + 2√6)^{(x2 - 3)}  

<sup>On putting these values of y and 1/y in equation in eq (1)

y + 1/y = 10
(y2 + 1)/y = 10
(y2 + 1) = 10y
y2 - 10y + 1 = 0

a = 1
b = -10
c = 1

D = b2 - 4ac
D = (-10)2 - 4(1)(1)
D = 100 - 4
D = 96 

Using Quadratic Formula

y = (- b ± √D) / 2a
y = [-(10) ± √96] / 2(1)
y = (10 ± 4√6) / 2 
y =  2(5 ± 2√6) / 2
y = 5 ± 2√6 

When y = 5  + 2√6</sup>

(5 + 2√6)(x²-3) = (5 + 2√6)¹
Equating the Power Both Sides
x² - 3 = 1
x² = 1 + 3
x² = 4 
x = ± √4
x = ± 2

When y = 5  - 2√6
(5 + 2√6)^(x2-3) = 5 - 2√6
(5 + 2√6)^(x2-3) = [1/(5 - 2√6)]^-1
(5 + 2√6)(x2-3) = (5 + 3√6)^-1
[Rationalising the denominator]
Equating the Power Both Sides
x² - 3 = -1
x² = -1+3|
x² = 2
x = ± √2

∴ x = ± 2, ±√2