Please simplify (2+root(3))= 1/((2-root(3)))Consider1/((2-root(3)))^x = tSolve the quadratic and then find x

The solutions to my question is not detailed, i need full details. Thanks

To tackle the problem you've presented, we first need to simplify the expression \( (2 + \sqrt{3}) = \frac{1}{(2 - \sqrt{3})} \). After that, we will consider the equation \( \frac{1}{(2 - \sqrt{3})^x} = t \) and solve for \( x \) by transforming it into a quadratic equation. Let’s break this down step by step.

Simplifying the Expression

We start with the expression \( (2 + \sqrt{3}) = \frac{1}{(2 - \sqrt{3})} \). To simplify this, we can multiply both sides by \( (2 - \sqrt{3}) \) to eliminate the fraction:

• First, we rewrite the equation: (2 + \sqrt{3})(2 - \sqrt{3}) = 1
• Next, we use the difference of squares formula: (a + b)(a - b) = a² - b²
• In our case, \( a = 2 \) and \( b = \sqrt{3} \), so: (2)² - (\sqrt{3})² = 4 - 3 = 1

This confirms that the original expression holds true, as both sides equal 1.

Finding x in the Equation

Next, let's move on to the equation \( \frac{1}{(2 - \sqrt{3})^x} = t \). We can rearrange this equation to isolate \( (2 - \sqrt{3})^x \):

• Taking the reciprocal gives us:(2 - \sqrt{3})^x = \frac{1}{t}

Now, we will take the logarithm of both sides to solve for \( x \):

• Applying logarithm:x \cdot \log(2 - \sqrt{3}) = \log\left(\frac{1}{t}\right)
• Using the property of logarithms:\log\left(\frac{1}{t}\right) = -\log(t)
• Now, we can express \( x \):x = \frac{-\log(t)}{\log(2 - \sqrt{3})}

Formulating a Quadratic Equation

To derive a quadratic equation, we can manipulate the expression for \( (2 - \sqrt{3})^x \). Let's rewrite it as \( y = (2 - \sqrt{3})^x \). Then we can substitute \( y \) back into our logarithmic equation:

• Rearranging gives us:y^x = \frac{1}{t}
• Substituting back:y = \frac{1}{t}

From this, we can derive a quadratic equation if we express \( y \) in terms of \( t \) and bring it to a standard form:

• Assuming \( y \) is a variable representing \( (2 - \sqrt{3})^x \):y^2 - k y + 1 = 0, where \( k \) is a constant derived from \( t \).

Solving the Quadratic Equation

Now, we can solve this quadratic equation using the quadratic formula:

• The quadratic formula is given by:y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
• Substituting \( a = 1 \), \( b = -k \), and \( c = 1 \):y = \frac{k \pm \sqrt{k^2 - 4}}{2}

This will yield the value of \( y \), which can then be substituted back to find the value of \( x \) using the relationship \( y = (2 - \sqrt{3})^x \).

Conclusion

By following these steps, we not only simplified the initial expression but also derived a way to express \( x \) in terms of logarithms and formulated a quadratic equation to solve for \( y \). This approach allows you to see the connections between algebraic manipulation and logarithmic expressions, which is essential in higher mathematics.