Question

The equation ${(6-x)}^4+{(8-x)}^4=16$ has

• A. Sum of roots $28$
• B. Product of roots $2688$
• C. Two real roots
• D. Two imaginary roots

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Sum of roots $28$
B Two real roots
C Product of roots $2688$
D Two imaginary roots

Given, ${(6-x)}^4+{(8-x)}^4=16\Rightarrow {(7-x-1)}^4+{(7-x+1)}^4=16$

Substituting $t=7-x$

${(t-1)}^4+{(t+1)}^4=16\Rightarrow t^4+6t^2-7=0\Rightarrow (t^2-1)(t^2+7)=0\Rightarrow t^2={(7-x)}^2=1,-7\Rightarrow 7-x=1,-1,\pm i\sqrt{7}\Rightarrow x=6,8,7+i\sqrt{7},7-i\sqrt{7}$

Hence, sum of roots is $S=28$

And product os roots is $P=48\times 56=2688$