Solve each of the following quadratic equations:

4^{(x + 1)} + 4^{(1 - x)} = 10

Given: 4^{(x + 1)} + 4^{(1 - x)} = 10

- - - - - - - (1)

Let 4^{x} = y - - - - - - - - - - (2)

substituting for y in (1)

4y^{2} - 10y + 4 = 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 = 4 b = - 10 c = 4

= 4.4 = 16

And either of their sum or difference = b

= - 10

Thus the two terms are - 8 and - 2

Sum = - 8 - 2 = - 10

Product = - 8. - 2 = 16

4y^{2} - 10y + 4 = 0

4y^{2} - 8y - 2y + 4 = 0

4y(y - 2) - 2(y - 2) = 0

(y - 2) (4y - 2) = 0

(y - 2) = 0 or (4y - 2) = 0

y = 2 or y = 1/2

substituting the value of y in (2)

4^{x} = 2 or 4^{x} = 2 ^{- 1}

2^{2x} = 2^{1} or 2^{2x} = 2 ^{- 1}

2x = 1 or 2x = - 1

Hence the roots of equation are

72