Understanding the Problem: Finding the Geometric Mean of Roots

The question asks for the Geometric Mean (GM) of the roots of a specific quadratic equation. A quadratic equation is given in the standard form \(Ax^2 + Bx + C = 0\), where \(A\), \(B\), and \(C\) are coefficients.

The given quadratic equation is:

\((3 + 2\sqrt{2})x^2 - (4 + 2\sqrt{3})x + (8 + 4\sqrt{3}) = 0\)

From this equation, we can identify the coefficients:

• \(A = 3 + 2\sqrt{2}\)
• \(B = -(4 + 2\sqrt{3})\)
• \(C = 8 + 4\sqrt{3}\)

Product of Roots of a Quadratic Equation

For a quadratic equation \(Ax^2 + Bx + C = 0\), if \(\alpha\) and \(\beta\) are the roots, the product of the roots is given by the formula:

\(\alpha \beta = \frac{C}{A}\)

Geometric Mean of Roots

The Geometric Mean (GM) of two numbers, \(\alpha\) and \(\beta\), is given by \(\sqrt{\alpha \beta}\) (assuming the roots are positive real numbers). Therefore, the GM of the roots of the quadratic equation is:

\(GM = \sqrt{\frac{C}{A}}\)

Calculating the Product of Roots \(C/A\)

Now, let's substitute the values of \(C\) and \(A\) into the formula for the product of roots:

\(\frac{C}{A} = \frac{8 + 4\sqrt{3}}{3 + 2\sqrt{2}}\)

We can simplify the expressions for \(A\) and \(C\) by recognizing them as perfect squares involving surds.

• For \(A = 3 + 2\sqrt{2}\): This is in the form \(a^2 + b^2 + 2ab\). If we consider \(a = \sqrt{2}\) and \(b = 1\), then \(a^2 = (\sqrt{2})^2 = 2\), \(b^2 = 1^2 = 1\), and \(2ab = 2(\sqrt{2})(1) = 2\sqrt{2}\). So, \(3 + 2\sqrt{2} = 2 + 1 + 2\sqrt{2} = (\sqrt{2} + 1)^2\).
• For \(C = 8 + 4\sqrt{3}\): We can factor out 4: \(4(2 + \sqrt{3})\). Now consider \(2 + \sqrt{3}\). If we evaluate \(\frac{(\sqrt{3}+1)^2}{2} = \frac{(\sqrt{3})^2 + 1^2 + 2\sqrt{3}}{2} = \frac{3+1+2\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}\). So, \(2 + \sqrt{3} = \frac{(\sqrt{3}+1)^2}{2}\). Therefore, \(C = 4 \times \frac{(\sqrt{3}+1)^2}{2} = 2(\sqrt{3}+1)^2\).

Now, substitute these simplified forms back into \(C/A\):

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

Calculating the Geometric Mean of the Roots

The GM is the square root of the product of roots:

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

Using the property \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\) and \(\sqrt{x^2} = |x|\):

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

Since \(\sqrt{3} + 1\) and \(\sqrt{2} + 1\) are both positive quantities, their absolute values are themselves:

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

To simplify this expression and match the options, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{2} - 1\):

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

Multiply the terms:

• Denominator: \((\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1\) (using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\)).
• Numerator: \(\sqrt{2} (\sqrt{3} + 1)(\sqrt{2} - 1)\)

First, multiply \((\sqrt{3} + 1)(\sqrt{2} - 1)\) using the distributive property (FOIL):

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

\(= \sqrt{6} - \sqrt{3} + \sqrt{2} - 1\)

Now, multiply this result by \(\sqrt{2}\):

\(\sqrt{2} (\sqrt{6} - \sqrt{3} + \sqrt{2} - 1)\)

Since the denominator is 1, the Geometric Mean is:

\(GM = \sqrt{2} (\sqrt{6} - \sqrt{3} + \sqrt{2} - 1)\)

Comparing with Options

Let's look at the given options and compare with our calculated GM:

• Option 1: \(\sqrt{2}(\sqrt{6} - \sqrt{3} + \sqrt{2} - 1)\)
• Option 2: \(\sqrt{2}(\sqrt{6} + \sqrt{3} - \sqrt{2} - 1)\)
• Option 3: \((\sqrt{6} - \sqrt{3} + \sqrt{2} - 1)\)
• Option 4: \((\sqrt{6} + \sqrt{3} + \sqrt{2} - 1)\)

Our calculated Geometric Mean, \(\sqrt{2} (\sqrt{6} - \sqrt{3} + \sqrt{2} - 1)\), exactly matches Option 1.

Summary of Calculation Steps

1. Identify the coefficients \(A\) and \(C\) from the given quadratic equation.
2. Recall the formula for the product of roots: \(\alpha \beta = C/A\).
3. Recall the formula for the Geometric Mean of roots: \(GM = \sqrt{\alpha \beta}\).
4. Substitute \(A\) and \(C\), simplifying them by recognizing them as perfect squares of surd expressions.
5. Calculate \(C/A\).
6. Take the square root of \(C/A\) to find the GM.
7. Simplify the resulting surd expression, including rationalizing the denominator.
8. Compare the final simplified expression for the GM with the given options to find the match.

Revision Table: Concepts for Quadratic Roots and Geometric Mean

Additional Information: Simplifying Surd Expressions

Working with quadratic equations where coefficients are surds requires careful manipulation of radical expressions. Two key techniques used in this problem were recognizing perfect square forms and rationalizing the denominator.

Recognizing Perfect Squares: Some expressions like \(3 + 2\sqrt{2}\) might look complex but can be simplified by recognizing the pattern of \((a+b)^2\) or \((\sqrt{a}+\sqrt{b})^2\). For \(3 + 2\sqrt{2}\), thinking of it as \(2 + 1 + 2\sqrt{2}\) helps see it as \((\sqrt{2})^2 + 1^2 + 2(\sqrt{2})(1) = (\sqrt{2} + 1)^2\).

Similarly, \(8 + 4\sqrt{3}\) can be written as \(4(2 + \sqrt{3})\). The term \(2 + \sqrt{3}\) is half of \(4 + 2\sqrt{3}\), which is \((\sqrt{3} + 1)^2\). So, \(2 + \sqrt{3} = \frac{(\sqrt{3} + 1)^2}{2}\), making \(8 + 4\sqrt{3} = 4 \times \frac{(\sqrt{3} + 1)^2}{2} = 2(\sqrt{3} + 1)^2\). This simplification was vital for easily calculating the square root of the product of roots.

Rationalizing the Denominator: When a fraction has a surd in the denominator, like \(\frac{P}{\sqrt{a} + \sqrt{b}}\) or \(\frac{P}{a + \sqrt{b}}\), we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{a} + \sqrt{b}\) is \(\sqrt{a} - \sqrt{b}\), and the conjugate of \(a + \sqrt{b}\) is \(a - \sqrt{b}\). This is because the product of a term and its conjugate results in a rational number (no surds), using the difference of squares formula \((x+y)(x-y) = x^2 - y^2\). In this problem, we rationalized \(\frac{\sqrt{2} (\sqrt{3} + 1)}{\sqrt{2} + 1}\) by multiplying by \(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\).