Given: 3^{(x + 2)} + 3 ^{- x} = 10

- - - - - - - - (1)

Let 3^x = y - - - - - - - - - - (2)

substituting for y in (1)

9y² - 10y + 1 = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 9 b = - 10 c = 1

= 9.1 = 9

And either of their sum or difference = b

= - 10

Thus the two terms are - 9 and - 1

Sum = - 9 - 1 = - 10

Product = - 9. - 1 = 9

9y² - 9y - 1y + 1 = 0

9y(y - 1) - 1(y - 1) = 0

(y - 1) (9y - 1) = 0

(y - 1) = 0 or (9y - 1) = 0

y = 1 or y = 1/9

3^x = 1 or 3^x = 1/9

On putting value of y in equation (2)

3^x = 3⁰ or 3^x = 3 ^{- 2}

x = 0 or x = - 2

Hence the roots of equation are 0, - 2