Question

$\text{Solve the equation}$
$(2+\sqrt{3} )^{x}+(2-\sqrt{3} )^{x}=4$

Answer 1

$\large{\text{\underline{Hints \& Methods:-}}}$

This is a simple exponential equation if we note the base of two powers is reciprocal.

Then we can build a quadratic equation for another variable, say $t$.

The method of solving the equation has three ways.

①Factorization.

②Complete the square.

③Quadratic formula.

Here we will use ③Quadratic formula.

$\large{\text{\underline{Step A: Reciprocal.}}}$

We are given that the sum of the two powers sum to 4. And the base of two numbers is reciprocal of each other.

$\hookrightarrow 2-\sqrt{3} =\dfrac{1}{2+\sqrt{3} }\text{.}$

Rebuilding the equation,

$\hookrightarrow (2+\sqrt{3} )^{x}+(\dfrac{1}{2+\sqrt{3} } )^{x}=4$

$\hookrightarrow (2+\sqrt{3} )^{x}+(2+\sqrt{3} )^{-x}=4\text{.}$

$\large{\text{\underline{Step B: Building an equation.}}}$

Now we can see common parts. For this case, we are going to set a new variable $t=(2+\sqrt{3})^{x}$.

Now,

$\hookrightarrow t+\dfrac{1}{t} =4$

$\hookrightarrow t^{2}-4t+1=0$

$\hookrightarrow t=2\pm\sqrt{3}$

$\large{\text{\underline{Step C: Solving the equation.}}}$

Now, substituting $t=(2+\sqrt{3})^{x}$ back in,

$\hookrightarrow (2+\sqrt{3})^{x}=2+\sqrt{3}\text{ or }(2+\sqrt{3})^{x}=2-\sqrt{3}$

$\hookrightarrow x=1\text{ or }x=-1\text{.}$

$\large{\text{\underline{Final answer.}}}$

Hence the solutions are $x=\pm1$.

Answer 2

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