Question

Solve the following quadratic equation:

`2^(2x) - 3·2^((x + 2)) + 32 = 0`

Answer 1

`2^(2x) - 3·2^((x + 2)) + 32 = 0` 

⇒ `(2^x)^2 - 3·2^x·2^2 + 32 = 0` 

Let 2^x be y. 

∴ y² – 12y + 32 = 0 

⇒ y² – 8y – 4y + 32 = 0 

⇒ y(y – 8) – 4(y – 8) = 0 

⇒ (y – 8) = 0 or (y – 4) = 0 

⇒ y = 8 or y = 4 

∴ 2^x = 8 or 2^x = 4 

⇒ 2^x = 2³ or 2^x = 2² 

⇒ x – 2 or 3  

Hence, 2 and 3 are the roots of the given equation.