Simplifying the Square Root Expression for x

The problem asks us to first find the value of \(x\) defined by a difference of two square roots, and then use this value to evaluate another algebraic expression.

The given expression for \(x\) is:

\( x = \sqrt {1 + \frac{{\sqrt 3 }}{2}} - \sqrt {1 - \frac{{{\kern 1pt} \sqrt 3 }}{2}} \)

We can simplify the terms inside the square roots by trying to express them in the form \( (a \pm b)^2 \). Recall that \( (a \pm b)^2 = a^2 + b^2 \pm 2ab \). Our goal is to make the terms look like \( a^2 + b^2 \pm 2ab \).

Consider the first term under the square root: \( 1 + \frac{{\sqrt 3 }}{2} \). We can rewrite this as \( \frac{2 + \sqrt{3}}{2} \). To make the numerator a perfect square of the form \( (a+b)^2 \), we need a \(2ab\) term. We have \(\sqrt{3}\). If we consider \(\sqrt{3}\) as \(2ab\), then \(ab = \frac{\sqrt{3}}{2}\). This doesn't immediately suggest simple \(a\) and \(b\).

Let's try a different approach. Multiply the numerator and denominator by 2:

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

We can write \( 2 + \sqrt{3} \) as \( (\sqrt{a} + \sqrt{b})^2 = a+b+2\sqrt{ab} \). We want \( a+b=2 \) and \( 2\sqrt{ab}=\sqrt{3} \), which means \( \sqrt{ab} = \frac{\sqrt{3}}{2} \) or \( ab = \frac{3}{4} \). We need two numbers that sum to 2 and multiply to \(\frac{3}{4}\). These numbers are \(\frac{3}{2}\) and \(\frac{1}{2}\).

So, \( 2 + \sqrt{3} = \frac{3}{2} + \frac{1}{2} + 2\sqrt{\frac{3}{2} \times \frac{1}{2}} = \left(\sqrt{\frac{3}{2}}\right)^2 + \left(\sqrt{\frac{1}{2}}\right)^2 + 2\sqrt{\frac{3}{4}} = \left(\sqrt{\frac{3}{2}}\right)^2 + \left(\sqrt{\frac{1}{2}}\right)^2 + 2 \frac{\sqrt{3}}{2} \). This still seems complicated.

Let's try multiplying the numerator and denominator of \( 1 + \frac{{\sqrt 3 }}{2} \) by 2 in a different way:

\( 1 + \frac{{\sqrt 3 }}{2} = \frac{2 + \sqrt{3}}{2} = \frac{4 + 2\sqrt{3}}{4} \)

Now, the numerator \( 4 + 2\sqrt{3} \) is easier to handle. We need \( a^2 + b^2 = 4 \) and \( 2ab = 2\sqrt{3} \), which means \( ab = \sqrt{3} \). Consider \( a=\sqrt{3} \) and \( b=1 \). Then \( a^2 = 3 \) and \( b^2 = 1 \). \( a^2 + b^2 = 3+1 = 4 \). This matches!

So, \( 4 + 2\sqrt{3} = (\sqrt{3})^2 + 1^2 + 2(\sqrt{3})(1) = (\sqrt{3}+1)^2 \).

Therefore, \( 1 + \frac{{\sqrt 3 }}{2} = \frac{(\sqrt{3}+1)^2}{4} \).

Similarly, for the second term under the square root:

\( 1 - \frac{{{\kern 1pt} \sqrt 3 }}{2} = \frac{2 - \sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{4} \)

The numerator \( 4 - 2\sqrt{3} \) is \( (\sqrt{3})^2 + 1^2 - 2(\sqrt{3})(1) = (\sqrt{3}-1)^2 \).

Therefore, \( 1 - \frac{{{\kern 1pt} \sqrt 3 }}{2} = \frac{(\sqrt{3}-1)^2}{4} \).

Calculating the Value of x

Now substitute these simplified forms back into the expression for \(x\):

\( x = \sqrt{\frac{(\sqrt{3}+1)^2}{4}} - \sqrt{\frac{(\sqrt{3}-1)^2}{4}} \)

\( x = \frac{\sqrt{(\sqrt{3}+1)^2}}{\sqrt{4}} - \frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{4}} \)

Using the property \( \sqrt{a^2} = |a| \):

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

Since \( \sqrt{3} \approx 1.732 \), we have \( \sqrt{3}+1 > 0 \) and \( \sqrt{3}-1 > 0 \).

So, \( |\sqrt{3}+1| = \sqrt{3}+1 \) and \( |\sqrt{3}-1| = \sqrt{3}-1 \).

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

Combine the fractions:

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

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

\( x = \frac{2}{2} \)

\( x = 1 \)

Evaluating the Expression \( \frac{{\sqrt 3 - x}}{{\sqrt 3 + x}} \)

Now that we have \( x=1 \), we can substitute this value into the expression we need to evaluate:

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

To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{3}-1 \).

Expression \( = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} \)

In the numerator, we have \( (\sqrt{3}-1)^2 = (\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \).

In the denominator, we have a difference of squares: \( (\sqrt{3}+1)(\sqrt{3}-1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2 \).

So, the expression becomes:

Expression \( = \frac{4 - 2\sqrt{3}}{2} \)

Divide both terms in the numerator by 2:

Expression \( = \frac{4}{2} - \frac{2\sqrt{3}}{2} \)

Expression \( = 2 - \sqrt{3} \)

Calculating the Numerical Value and Rounding

We need the value corrected to two decimal places. We use the approximate value of \( \sqrt{3} \).

Using \( \sqrt{3} \approx 1.73205 \):

Value \( = 2 - 1.73205 \)

Value \( = 0.26795 \)

Rounding this value to two decimal places:

• The first decimal place is 2.
• The second decimal place is 6.
• The third decimal place is 7.

Since the third decimal place (7) is 5 or greater, we round up the second decimal place.

\( 0.26795 \approx 0.27 \)

The value of \( \frac{{\sqrt 3 - x}}{{\sqrt 3 + x}} \) corrected to two decimal places is \(0.27\).

Revision Table: Key Concepts

Additional Information: Manipulating Square Roots and Surds

Manipulating expressions with square roots, also known as surds, is a fundamental skill in algebra. Here are some key techniques used:

• Combining Like Terms: Just like variables, surds with the same number under the root can be added or subtracted, e.g., \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\).
• Multiplying and Dividing Surds: \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\) and \(\sqrt{a} / \sqrt{b} = \sqrt{a/b}\).
• Simplifying Surds: Extracting perfect square factors from under the root, e.g., \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\).
• Rationalizing Denominators: Removing surds from the denominator of a fraction.
  • If the denominator is a single surd like \(\sqrt{a}\), multiply numerator and denominator by \(\sqrt{a}\).
  • If the denominator is a binomial involving surds like \(a + \sqrt{b}\) or \(\sqrt{a} + \sqrt{b}\), multiply numerator and denominator by its conjugate (\(a - \sqrt{b}\) or \(\sqrt{a} - \sqrt{b}\)). The product of a binomial and its conjugate results in a rational number: \((x+y)(x-y) = x^2 - y^2\).
• Recognizing Perfect Squares: Being able to spot expressions that are perfect squares, such as \(a \pm b \pm 2\sqrt{ab}\) which comes from \((\sqrt{a} \pm \sqrt{b})^2\), or in this problem, recognizing \(4 \pm 2\sqrt{3}\) as \((\sqrt{3} \pm 1)^2\). This often involves manipulating the expression to get a '2' in front of the surd term if it's intended to be part of a \(2ab\) term.

These techniques are essential for simplifying expressions and solving equations involving square roots.