Solve each of the following quadratic equations:

2^{2x} - 3.2^{(x + 2)} + 32 = 0

Given: 2^{2x} - 3.2^{(x + 2)} + 32 = 0

(2^{x})^{2} - 3.2^{x}.2^{2} + 32 = 0 - - - - (1)

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

substituting for y in (1)

y^{2} - 12y + 32 = 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 = 1 b = - 12 c = 32

= 1.32 = 32

And either of their sum or difference = b

= - 12

Thus the two terms are - 8 and - 4

Sum = - 8 - 4 = - 12

Product = - 8. - 4 = 32

y^{2} - 8y - 4y + 32 = 0

y(y - 8) - 4(y - 8) = 0

(y - 8) (y - 4) = 0

(y - 8) = 0 or (y - 4) = 0

y = 8 or y = 4

2^{x} = 8 or 2^{x} = 4

substituting the value of y in (2)

2^{x} = 2^{3} or 2^{x} = 2^{2}

x = 2 or x = 3

Hence the roots of equation are 2, 3

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