Question

If the equation ${(3-{\log }_{\sqrt[12]{124}}(x-2))}^2-4|3-{\log }_{\sqrt[12]{124}}(x-2)|+3=0$ has integral roots $\alpha$ and $\beta$, then $|\alpha +\beta |$ is equal to

• A. $7$
• B. $6$
• C. $5$
• D. $3$

Answer 1

The correct option is A $7$

${(3-{\log }_{\sqrt[12]{124}}(x-2))}^2-4|3-{\log }_{\sqrt[12]{124}}(x-2)|+3=0$
For the $\log$ to be defined,
$x>2$

Assuming $3-{\log }_{\sqrt[12]{124}}(x-2)=t$, we get
$t^2-4|t|+3=0\Rightarrow (|t|{)}^2-4|t|+3=0\Rightarrow (|t|-1\left)\right(|t|-3)=0\Rightarrow |t|=1,3$

Now,
$|3-{\log }_{\sqrt[12]{124}}(x-2)|=1\Rightarrow 3-{\log }_{\sqrt[12]{124}}(x-2)=\pm 1\Rightarrow -{\log }_{\sqrt[12]{124}}(x-2)=-3\pm 1\Rightarrow {\log }_{\sqrt[12]{124}}(x-2)=2,4\Rightarrow x-2=(\sqrt[12]{124}{)}^2,(\sqrt[12]{124}{)}^4\Rightarrow x=2+2^{1/3},2+2^{2/3}$
Which are not integers.

Also,
$|3-{\log }_{\sqrt[12]{124}}(x-2)|=3\Rightarrow 3-{\log }_{\sqrt[12]{124}}(x-2)=\pm 3$
$\Rightarrow {\log }_{\sqrt[12]{124}}(x-2)=0,6\Rightarrow x-2=(\sqrt[12]{124}{)}^0,(\sqrt[12]{124}{)}^6\Rightarrow x-2=1,2\Rightarrow x=3,4$

Hence, $|\alpha +\beta |=7$