Question

The number of real roots of ${(6-x)}^4+{(8-x)}^4=16$ is

• A. $0$
• B. $4$
• C. $2$
• D. None of these

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Find the roots of the given equation.

An equation ${(6-x)}^4+{(8-x)}^4=16$ is given.

Assume that,

$$\begin{gathered} y=\frac{\left(6-x\right)+(8-x)}{2} \\ \Rightarrow y=7-x \\ \Rightarrow x=7-y \end{gathered}$$

So, the given equation becomes:

$$\begin{gathered} {(6-7+y)}^4+{(8-7+y)}^4=16 \\ \Rightarrow {\left(y-1\right)}^4+{\left(y+1\right)}^4=16 \\ \Rightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=16 \\ \Rightarrow 2y^4+12y^2+2=16 \\ \Rightarrow y^4+6y^2+1=8 \\ \Rightarrow y^4+6y^2-7=0 \\ \Rightarrow (y^2-1)(y^2+7)=0 \end{gathered}$$

Since, $y^2+7\ne 0$ therefore, $y^2-1=0$.

So, $y=\pm 1$

Therefore, $x=6,8$.

Thus, the number of real roots of ${(6-x)}^4+{(8-x)}^4=16$ is $2$.

Hence, option $C$ is the correct option.