Understanding the Equation and its Roots

The given equation is: \({\left( {{x^2} + 2} \right)^2} + 8{x^2} = 6x\left( {{x^2} + 2} \right)\).

We need to analyze the nature of its roots and their sum. Let's rearrange and simplify the equation to identify its type and find its roots.

First, move all terms to one side to set the equation to zero:

\({\left( {{x^2} + 2} \right)^2} - 6x\left( {{x^2} + 2 \right) + 8{x^2} = 0\)

This equation looks like a quadratic expression if we consider \({{x^2} + 2}\) as one variable. Let's treat it that way for factoring.

We are looking for two terms that multiply to \(8x^2\) and add up to \(-6x\), when combined with the \({{x^2} + 2}\) terms. Let \(Y = x^2+2\). The equation becomes \(Y^2 - 6xY + 8x^2 = 0\). This is a quadratic in Y. We can factor this as \((Y - a x)(Y - b x) = 0\). We need \((-ax) + (-bx) = -6x\) and \((-ax)(-bx) = 8x^2\). This gives \(a+b = 6\) and \(ab = 8\). The numbers 2 and 4 satisfy these conditions.

So, the equation can be factored as:

\(\left( {\left( {{x^2} + 2} \right) - 2x} \right)\left( {\left( {{x^2} + 2} \right) - 4x} \right) = 0\)

This simplifies to:

\(\left( {{x^2} - 2x + 2} \right)\left( {{x^2} - 4x + 2} \right) = 0\)

For the entire expression to be zero, at least one of the factors must be zero. This gives us two separate quadratic equations to solve for the roots of the original equation:

1. \({{x^2} - 2x + 2 = 0}\)
2. \({{x^2} - 4x + 2 = 0}\)

Finding the Roots of the Equation

We will use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), to find the roots of each quadratic equation.

Equation 1: \(x^2 - 2x + 2 = 0\)

Here, \(a=1\), \(b=-2\), \(c=2\).

The discriminant is \(\Delta_1 = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4\).

Since \(\Delta_1 < 0\), the roots are complex conjugates.

The roots are \(x = \frac{-(-2) \pm \sqrt{-4}}{2(1)} = \frac{2 \pm 2i}{2} = 1 \pm i\).

The roots from this equation are \(1+i\) and \(1-i\).

Equation 2: \(x^2 - 4x + 2 = 0\)

Here, \(a=1\), \(b=-4\), \(c=2\).

The discriminant is \(\Delta_2 = b^2 - 4ac = (-4)^2 - 4(1)(2) = 16 - 8 = 8\).

Since \(\Delta_2 > 0\), the roots are real and distinct.

The roots are \(x = \frac{-(-4) \pm \sqrt{8}}{2(1)} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}\).

The roots from this equation are \(2+\sqrt{2}\) and \(2-\sqrt{2}\).

The four roots of the original quartic equation \({\left( {{x^2} + 2} \right)^2} + 8{x^2} = 6x\left( {{x^2} + 2 \right)}\) are \(1+i\), \(1-i\), \(2+\sqrt{2}\), and \(2-\sqrt{2}\).

Analyzing the Given Statements

Statement 1: All the roots of the equation are complex.

The roots we found are \(1+i\), \(1-i\) (complex) and \(2+\sqrt{2}\), \(2-\sqrt{2}\) (real). Since two of the roots are real numbers, the statement that all the roots are complex is incorrect.

Statement 1 is False.

Statement 2: The sum of all the roots of the equation is 6.

Let's calculate the sum of the four roots:

Sum = \((1+i) + (1-i) + (2+\sqrt{2}) + (2-\sqrt{2})\)

Sum = \(1 + i + 1 - i + 2 + \sqrt{2} + 2 - \sqrt{2}\)

Sum = \((1+1+2+2) + (i-i) + (\sqrt{2}-\sqrt{2})\)

Sum = \(6 + 0 + 0 = 6\)

Alternatively, we can verify the sum of roots using Vieta's formulas. The expanded form of the equation is \(x^4 - 6x^3 + 12x^2 - 12x + 4 = 0\). For a polynomial \(ax^n + bx^{n-1} + \dots + k = 0\), the sum of the roots is given by \(-b/a\).

In our equation \(x^4 - 6x^3 + 12x^2 - 12x + 4 = 0\), \(a=1\) and \(b=-6\). The sum of the roots is \(-(-6)/1 = 6\).

Both methods confirm that the sum of all the roots is 6.

Statement 2 is True.

Conclusion on Correct Statements

Based on our analysis:

• Statement 1 is incorrect.
• Statement 2 is correct.

Therefore, only Statement 2 is correct.

Revision Table: Equation Roots Analysis

To summarize the key findings related to the equation and its roots:

• Equation: \({\left( {{x^2} + 2} \right)^2} + 8{x^2} = 6x\left( {{x^2} + 2 \right)}\)
• Factored form: \(\left( {{x^2} - 2x + 2} \right)\left( {{x^2} - 4x + 2} \right) = 0\)
• Roots from \({{x^2} - 2x + 2 = 0}\): \(1+i, 1-i\) (complex)
• Roots from \({{x^2} - 4x + 2 = 0}\): \(2+\sqrt{2}, 2-\sqrt{2}\) (real)
• Total roots: \(1+i, 1-i, 2+\sqrt{2}, 2-\sqrt{2}\)
• Nature of roots: Two complex, two real. Not all roots are complex.
• Sum of roots: \((1+i)+(1-i)+(2+\sqrt{2})+(2-\sqrt{2}) = 6\)

Additional Information: Complex Roots and Vieta's Formulas

Complex Roots

Complex numbers are numbers of the form \(a+bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with \(i^2 = -1\). If a polynomial equation with real coefficients has a complex root \(a+bi\) where \(b \neq 0\), then its conjugate \(a-bi\) must also be a root. This is known as the Complex Conjugate Root Theorem. In our case, \(1+i\) is a root, and its conjugate \(1-i\) is also a root, as expected since the original equation has real coefficients.

Vieta's Formulas

Vieta's formulas provide relationships between the roots of a polynomial equation and its coefficients. For a polynomial equation of degree \(n\), \(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0\), with roots \(r_1, r_2, \dots, r_n\), the sum of the roots is given by \(r_1 + r_2 + \dots + r_n = -\frac{a_{n-1}}{a_n}\). For our quartic equation \(x^4 - 6x^3 + 12x^2 - 12x + 4 = 0\), the coefficient of \(x^4\) is \(a_4=1\) and the coefficient of \(x^3\) is \(a_3=-6\). The sum of the roots is indeed \(-\frac{-6}{1} = 6\), confirming our calculation.