Question

solve the equation
${(2+\sqrt{3})}^{x^2-2x+1}+{(2-\sqrt{3})}^{x^2-2x-1}=\frac{101}{10(2-\sqrt{3})}$
find the sum of the solutions of the equation.

Answer 1

${(2+\sqrt{3})}^{x^2-2x+1}+{(2-\sqrt{3})}^{x^2-2x-1}=\frac{101}{10(2-\sqrt{3})}$

$(2+\sqrt{3})=\frac{1}{(2-\sqrt{3})}$

So we can write the above equation
as

${(2+\sqrt{3})}^{x^2-2x+1}+{\left(\frac{1}{2+\sqrt{3}}\right)}^{x^2-2x-1}=\frac{101(2+\sqrt{3})}{10}$

Now Let us assume ${(2+\sqrt{3})}^{x^2-2x}=t$

Then we reduce the equation
to

$(2+\sqrt{3})t+\frac{(2+\sqrt{3})}{t}=\frac{101(2+\sqrt{3})}{10}$

So we get

$t+\frac{1}{t}=10+\frac{1}{10}$

So $t=10,\frac{1}{10}$

${(2+\sqrt{3})}^{x^2-2x}=10$ and ${(2+\sqrt{3})}^{x^2-2x}=\frac{1}{10}$

Taking Log on both sides we get

$x^2-2x-{\log }_{2+\sqrt{3}}10=0$ and

$x^2-2x+{\log }_{2+\sqrt{3}}10=0$

So Sum of the Roots for both the equation
is $2$ and $2$

So the Sum of the Solution
of the equation
is $4$