Question

The number of real roots of $(3-x{)}^4+(5-x{)}^4=16$ is

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

Answer 1

The correct option is B $2$

$(3-x{)}^4+(5-x{)}^4=16$

Now, put $4-x=t$

$\Rightarrow$ $(t-1{)}^4+(t+1{)}^4=16$

$\Rightarrow$ $(t^4-4t^3+6t^2-4t+1)+(t^4+4t^3+6t^2+4t+1)=16$

$\Rightarrow$ $2(t^4+6t^2+1)=16$

$\Rightarrow$ $t^4+6t^2+1=8$

Now, put $t^2=p$, so we get

$\Rightarrow$ $p^2+6p-7=0$

$\Rightarrow$ $(p+7)(p-1)=0$

$\therefore$ $p=1$ and $p=-7$

Now, $p=-7$ will not yield any real solution so, we ignore that

$\Rightarrow$ $t^2=1$

$\therefore$ $t=\pm 1$

Substituting $t=1$

$4-x=1$

$\therefore$ $x=3$

Put $t=-1$

$4-x=-1$

$\therefore$ $x=5$

$\therefore$ We get, $x=3,5$

Hence, only $two$ real roots we will get.