Question

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

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Let, $7-x=a$
The equation becomes
$(a-1{)}^4+(a+1{)}^4=16$
$\Rightarrow a^4+6a^2+1=8$
$\Rightarrow a^4+6a^2-7=0$
$\Rightarrow (a^2+7)(a^2-1)=0$
Now, $a^2>0$,
So, $a^2=1$
Hence, $a=1,a=-1$
$\Rightarrow x=6$ or $x=8$.
Hence, there are two real solutions of $x$.