Real Roots And Cubic Equations Exploring Solutions And Relationships

In the realm of mathematical problem-solving, equations often present themselves as intricate puzzles, demanding a keen understanding of algebraic principles and a strategic approach to unravel their solutions. One such equation that captures our attention is:

${\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}}$

Our mission is to meticulously dissect this equation, identify the constraints governing its domain, and ultimately determine the number of real roots that satisfy its enigmatic nature. In this comprehensive analysis, we will embark on a step-by-step journey, employing a combination of algebraic manipulation, domain considerations, and graphical insights to arrive at the definitive answer.

17.1 The Quest for Real Roots

At the heart of our investigation lies the pursuit of real roots – those tangible values of 'x' that, when plugged into the equation, render it a harmonious truth. However, the presence of square roots introduces a subtle yet crucial layer of complexity. The expression under each square root must be non-negative to yield real values, imposing constraints on the permissible domain of 'x'.

17.1.1 Domain Deliberations: Unveiling the Boundaries

Let's embark on a meticulous examination of the domain restrictions imposed by each square root term:

1. ${\sqrt{x^2 - 4x + 3}}$: The expression ${x^2 - 4x + 3}$ must be greater than or equal to zero. Factoring this quadratic, we get ${(x - 1)(x - 3) \ge 0}$. This inequality holds true when ${x \le 1}$ or ${x \ge 3}$.
2. ${\sqrt{x^2 - 9}}$: Similarly, ${x^2 - 9}$ must be non-negative. Factoring this difference of squares, we get ${(x - 3)(x + 3) \ge 0}$. This inequality is satisfied when ${x \le -3}$ or ${x \ge 3}$.
3. ${\sqrt{4x^2 - 14x + 6}}$: The expression ${4x^2 - 14x + 6}$ must also be non-negative. We can simplify this by dividing by 2, yielding ${2x^2 - 7x + 3 \ge 0}$. Factoring this quadratic, we obtain ${(2x - 1)(x - 3) \ge 0}$. This inequality holds when ${x \le \frac{1}{2}}$ or ${x \ge 3}$.

17.1.2 The Interplay of Domains: A Unified Perspective

To satisfy the original equation, all three square root terms must be real. Therefore, we need to find the intersection of the domains derived above. By carefully considering the intervals where each inequality holds, we arrive at the following permissible domains for 'x':

• ${x \le -3}$
• ${x = 3}$

17.1.3 Squaring and Simplifying: A Path to Potential Solutions

To eliminate the square roots and simplify the equation, we square both sides. This operation, while powerful, introduces a caveat: it may also introduce extraneous solutions – values that satisfy the transformed equation but not the original. Therefore, it is crucial to verify any solutions obtained in the original equation.

Squaring both sides of the original equation, we get:

${(\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9})^2 = (\sqrt{4x^2 - 14x + 6})^2}$

Expanding the left side, we have:

${(x^2 - 4x + 3) + 2\sqrt{(x^2 - 4x + 3)(x^2 - 9)} + (x^2 - 9) = 4x^2 - 14x + 6}$

Simplifying and rearranging terms, we obtain:

${2\sqrt{(x^2 - 4x + 3)(x^2 - 9)} = 2x^2 - 10x + 12}$

Dividing both sides by 2, we get:

${\sqrt{(x^2 - 4x + 3)(x^2 - 9)} = x^2 - 5x + 6}$

Squaring both sides again to eliminate the remaining square root:

${(x^2 - 4x + 3)(x^2 - 9) = (x^2 - 5x + 6)^2}$

Expanding and simplifying, we arrive at:

${x^4 - 4x^3 - 6x^2 + 36x - 27 = x^4 - 10x^3 + 37x^2 - 60x + 36}$

Further simplification leads to:

${6x^3 - 43x^2 + 96x - 63 = 0}$

17.1.4 Root Revelation: Identifying the Solutions

We now have a cubic equation to solve. While general cubic equations can be solved using Cardano's method, in this case, we can attempt to find rational roots using the Rational Root Theorem. This theorem suggests that any rational root of the equation must be a divisor of the constant term (-63) divided by a divisor of the leading coefficient (6).

By testing potential rational roots, we discover that ${x = 3}$ is indeed a root of the cubic equation. Performing polynomial division, we can factor the cubic equation as follows:

${(x - 3)(6x^2 - 25x + 21) = 0}$

Now we need to solve the quadratic equation ${6x^2 - 25x + 21 = 0}$. Factoring this quadratic, we get:

${(2x - 3)(3x - 7) = 0}$

This yields two additional potential roots: ${x = \frac{3}{2}}$ and ${x = \frac{7}{3}}$.

17.1.5 The Verdict: Validation and Extraneous Roots

We have identified three potential roots: ${x = 3}$, ${x = \frac{3}{2}}$, and ${x = \frac{7}{3}}$. However, we must now verify these solutions in the original equation to eliminate any extraneous roots.

1. ${x = 3}$: Substituting ${x = 3}$ into the original equation, we get:

   ${\sqrt{3^2 - 4(3) + 3} + \sqrt{3^2 - 9} = \sqrt{4(3)^2 - 14(3) + 6}}$

   ${\sqrt{0} + \sqrt{0} = \sqrt{0}}$

   This holds true, so ${x = 3}$ is a valid root.

2. ${x = \frac{3}{2}}$: This value is not within the permissible domain (${x \le -3}$ or ${x \ge 3}$), so it is an extraneous root.

3. ${x = \frac{7}{3}}$: This value is also not within the permissible domain (${x \le -3}$ or ${x \ge 3}$), so it is an extraneous root.

Therefore, the only real root of the equation is ${x = 3}$.

17.2 Conclusion: The Singular Solution

Through a rigorous analysis of the equation ${\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}}$, we have determined that it possesses only one real root, namely ${x = 3}$. This journey involved careful consideration of domain restrictions, algebraic manipulation to eliminate square roots, and verification of potential solutions to identify and discard extraneous roots. The process exemplifies the importance of a systematic approach when tackling complex mathematical problems.

In the captivating world of algebra, cubic equations hold a special allure. Their three roots, often intertwined in intricate relationships, offer a gateway to deeper mathematical insights. Let's delve into the realm of cubic equations, focusing on the equation:

${x^3 + bx + c = 0}$

where ${\alpha}$, ${\beta}$, and ${\gamma}$ are the three roots. Our investigation will center on a specific condition: ${\beta\gamma = 1}$. Our goal is to unravel the implications of this condition, to forge connections between the coefficients ('b' and 'c') and the roots themselves.

18.1 The Symphony of Roots Unveiling Vieta's Formulas

Cubic equations, like their quadratic cousins, adhere to a set of fundamental relationships between their roots and coefficients. These relationships, elegantly encapsulated in Vieta's formulas, serve as our guiding principles in this exploration. For the cubic equation ${x^3 + bx + c = 0}$, Vieta's formulas state:

1. Sum of Roots: ${\alpha + \beta + \gamma = 0}$
2. Sum of Pairwise Products: ${\alpha\beta + \alpha\gamma + \beta\gamma = b}$
3. Product of Roots: ${\alpha\beta\gamma = -c}$

These formulas provide a powerful lens through which to examine the interplay between the roots and the coefficients. In our specific case, the condition ${\beta\gamma = 1}$ adds a unique flavor to this interplay.

18.1.1 Decoding the Condition: ${\beta\gamma = 1}$

The condition ${\beta\gamma = 1}$ suggests a reciprocal relationship between the roots ${\beta}$ and ${\gamma}$. This hints at a certain symmetry or balance within the solution set. To fully grasp the implications, let's weave this condition into the fabric of Vieta's formulas.

Substituting ${\beta\gamma = 1}$ into the third Vieta's formula, we obtain:

${\alpha(1) = -c}$

This elegantly reveals that:

${\alpha = -c}$

This is a significant breakthrough. We have expressed one of the roots, ${\alpha}$, directly in terms of the coefficient 'c'. This connection paves the way for further deductions.

18.1.2 The Dance of Coefficients: Expressing 'b' and 'c'

Now, let's leverage the first two Vieta's formulas, armed with our newfound knowledge of ${\alpha}$. Substituting ${\alpha = -c}$ and ${\beta\gamma = 1}$ into the first Vieta's formula, we get:

${-c + \beta + \gamma = 0}$

This implies that:

${\beta + \gamma = c}$

This equation links the sum of the roots ${\beta}$ and ${\gamma}$ directly to the coefficient 'c'. Next, let's consider the second Vieta's formula:

${\alpha\beta + \alpha\gamma + \beta\gamma = b}$

Substituting ${\alpha = -c}$ and ${\beta\gamma = 1}$, we get:

${(-c)\beta + (-c)\gamma + 1 = b}$

Factoring out '-c', we have:

${-c(\beta + \gamma) + 1 = b}$

We already know that ${\beta + \gamma = c}$, so substituting this in, we obtain:

${-c(c) + 1 = b}$

This simplifies to:

${b = 1 - c^2}$

We have now expressed 'b' in terms of 'c'. This marks a significant step towards understanding the relationship between the coefficients and the roots.

18.1.3 Unveiling the Equation: A Coherent Narrative

We have successfully established a connection between the coefficients 'b' and 'c', encapsulated in the equation:

${b = 1 - c^2}$

This equation represents a constraint that must be satisfied given the condition ${\beta\gamma = 1}$. This equation is the key to solving the problem.

18.2 Conclusion: The Harmony of Roots and Coefficients

Through a systematic application of Vieta's formulas and a keen focus on the condition ${\beta\gamma = 1}$, we have unveiled a fundamental relationship between the coefficients 'b' and 'c' in the cubic equation ${x^3 + bx + c = 0}$. The equation ${b = 1 - c^2}$ stands as a testament to the interconnectedness of roots and coefficients in polynomial equations. This exploration highlights the power of Vieta's formulas as a tool for unraveling the secrets hidden within algebraic structures.

In this comprehensive analysis, we've journeyed through the landscape of real roots and cubic equations, employing algebraic manipulation, domain considerations, and the elegant power of Vieta's formulas. These mathematical tools serve as invaluable guides in the quest for solutions, illuminating the intricate relationships that govern the world of equations. By understanding the fundamental concepts, we empower ourselves to tackle complex problems and uncover the beauty of mathematical truths.