Question

Sum of the roots of an equation, $4^x-3\left(2^{x+3}\right)+128=0$, is:

• A. $0$
• B. $7$
• C. $5$
• D. $8$

Answer 1

The correct option is B

$7$

Find the roots of the equation by the factorization method:

It is given that $4^x-3\left(2^{x+3}\right)+128=0$,

Simplify the above equation as:

$\begin{array}{rcl}& \Rightarrow & {\left(2^x\right)}^2-3\left(2^x{2}^3\right)+128=0\left[\because a^{m+n}=a^m·a^n\right]\\ & \Rightarrow & {\left(2^x\right)}^2-24\left(2^x\right)+128=0\end{array}$

Let $y=2^x$, then the above equation becomes

$y^2-24y+128=0$

Factorize the above equation by splitting the middle term as follows:

$$\begin{gathered} y^2-16y-8y+128=0 \\ \Rightarrow y\left(y-16\right)-8\left(y-16\right)=0 \\ \Rightarrow \left(y-8\right)\left(y-16\right)=0 \\ \Rightarrow y=8,16 \end{gathered}$$

Substitute back $y=2^x$ and simplify further.

$$\begin{gathered} 2^x=8 \\ \Rightarrow 2^x=2^3 \\ \Rightarrow x=3 \end{gathered}$$

and,

$$\begin{gathered} 2^x=16 \\ \Rightarrow 2^x=2^4 \\ \Rightarrow x=4 \end{gathered}$$

Here, the roots are $3$ and $4$.

Therefore, the sum of the roots is $3+4=7$.

Hence, the correct option is (B).