Question

What is the sum of the roots of the equation ${{4}^{x}}-3\left( {{2}^{x+3}} \right)+128=0$ ?

Answer 1

Hint: We can find the sum of the roots of an equation by finding all the roots first. We can use some conversion methods to convert the given equation into a quadratic equation so that we can find its roots easily. Once we do, we can find its sum.

Complete step by step answer:
In the question we are given the equation, ${{4}^{x}}-3\left( {{2}^{x+3}} \right)+128=0$ . We can write this equation also as,
$\begin{align}
  & {{\left( {{2}^{2}} \right)}^{x}}-3\left( {{2}^{x}}{{2}^{3}} \right)+128=0 \\
 & \,\,{{\left( {{2}^{x}} \right)}^{2}}-24\left( {{2}^{x}} \right)+128=0 \\
\end{align}$
This equation is not in polynomial form, hence it would be difficult to find its roots directly. So, we will convert it into a quadratic equation by substituting ${{2}^{x}}$ as let’s say y. Doing so, we get,
$\,\,{{y}^{2}}-24y+128=0$
Now, we can clearly see that this is a quadratic equation with y as the variable. Let us proceed to solve this equation by finding factors of 128 and expressing 24 as a sum of its factors.
$128=16\times 8$
Hence, we can express 24 as the sum of 16 and 8 and then find the roots of the quadratic equation. We get,
$\begin{align}
  & \,\,\,\,{{y}^{2}}-16y-8y+128=0 \\
 & y\left( y-16 \right)-8\left( y-16 \right)=0 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\left( y-16 \right)\left( y-8 \right)=0 \\
\end{align}$
Hence we have found the roots as, $y=8,16$ . To find the roots of the original equation, we will back substitute y as x and find the roots in terms of x.
$\begin{align}
  & \,\,\,\,{{2}^{x}}=8 \\
 & \,\,\,\,{{2}^{x}}={{2}^{3}} \\
 & \Rightarrow x=3 \\
\end{align}$
$\begin{align}
  & \,\,\,\,{{2}^{x}}=16 \\
 & \,\,\,\,{{2}^{x}}={{2}^{4}} \\
 & \Rightarrow x=4 \\
\end{align}$
Therefore, the roots of the given equation are 3 and 4. The sum of the roots would then be 7 which is the required result.

Note: In this question, when we solve for the quadratic, we have to be careful while taking the factors. Since 32 and 8 are factors of 128 and also give a difference of 24, we can write the quadratic equation as,
$\,\,\,\,{{y}^{2}}-32y+8y+128=0$
But, in the next step we can clearly see that we cannot obtain roots from this.
$\,\,\,\,y\left( y-32 \right)+8\left( y+6 \right)=0$
Hence, we must take the right factors which would enable us to find the roots.