Question

Solve the equation:- \[{{2}^{2x}}+{{2}^{x+2}}-4\times {{2}^{3}}=0\].
(A) \[({{2}^{x}}-4)({{2}^{x}}+8)\]
(B) \[({{2}^{x}}-4)({{2}^{x}}-8)\]
(C) \[({{2}^{x}}+4)({{2}^{x}}+8)\]
(D) \[({{2}^{x}}+4)({{2}^{x}}-8)\]

Answer 1

Hint: In the given question, all the numbers are having the same base, that is base two, but their powers are different, and using the rules of multiplication and addition of the algebraic expression we can solve the equation. We will consider the value of two to the power x as ‘t’ and then solving the equation will be easy.

Complete step by step answer:
In mathematics, we have the power of numbers. The power of a number represents how many times the number will be multiplied in the question. We also have the power of numbers in the form of algebraic expressions. If we have two same numbers but their powers are different, and those numbers are multiplied by each other then the result will be the same number but their powers will be added to each other. If we have two numbers with the same base but they are having different powers and they both are divided by each other then the result will be the same number but their powers will be subtracted.
we have to solve the equation \[{{2}^{2x}}+{{2}^{x+2}}-4\times {{2}^{3}}=0\].
\[\begin{align}
  & {{2}^{2x}}+{{2}^{x+2}}-4\times {{2}^{3}}=0 \\
 & \Rightarrow {{({{2}^{x}})}^{2}}+{{2}^{x}}{{2}^{2}}-32=0 \\
\end{align}\]
Let the value of \[{{2}^{x}}=t\]. On putting this value in the above equation we get the following results.
\[{{t}^{2}}-4t-32=0\]……eq(1)
Now, this is in the form of the quadratic equation. We will solve this equation and will find out the value of ‘t’.
To solve the quadratic equation, first, we will do the factors of \[32\].
\[32=2\times 2\times 2\times 2\times 2\]
Now by using these factors, we will make \[4\]. So the difference of eight and four gives us four and when we multiply eight and four then we get \[32\].
So on putting this value in eq(1), the following results will be obtained which are as shown below.
\[\begin{align}
  & {{t}^{2}}-4t-32=0 \\
 & \Rightarrow {{t}^{2}}-(8-4)t-32=0 \\
 & \Rightarrow {{t}^{2}}-8t+4t-32=0 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow t(t-8)+4(t-8)=0 \\
 & \Rightarrow (t-8)(t+4)=0 \\
\end{align}\]
In the above equations, we have considered the value of \[{{2}^{x}}=t\]. Again putting the original value in the above equation, we get the following result.
\[{{2}^{2x}}+{{2}^{x+2}}-4\times {{2}^{3}}=({{2}^{x}}+4)({{2}^{x}}-8)\]

So, the correct answer is “Option D”.

Note:
 Every number in mathematics will have some power on it. A single number like seven will have power one and if any number will have power equals zero then the result will always be one. The number can have both positive and negative powers. If the number has negative power then after doing its reciprocal, its power will become positive.