For What Value Of A Does

For What Value of a Does the Equation x² + ax + a = 0 Have Exactly One Real Solution?

Finding the value of 'a' for which the quadratic equation x² + ax + a = 0 possesses only one real solution is a problem that elegantly blends algebra and the understanding of quadratic functions. This seemingly simple equation holds a wealth of mathematical concepts, from the discriminant to the nature of roots and the graphical representation of parabolas. Let's explore this problem step-by-step, ensuring a thorough understanding of the underlying principles.

Understanding Quadratic Equations and Their Solutions

Before delving into the specifics of our problem, it's crucial to establish a solid foundation in understanding quadratic equations. A quadratic equation is any equation that can be written in the standard form:

ax² + bx + c = 0

where a, b, and c are constants, and a ≠ 0. The solutions (or roots) of this equation represent the x-values where the parabola defined by the function y = ax² + bx + c intersects the x-axis (i.e., where y = 0).

The nature of these solutions—whether they are real, distinct, or equal—is determined by the discriminant, denoted as Δ (Delta):

Δ = b² - 4ac

• If Δ > 0: The equation has two distinct real solutions. The parabola intersects the x-axis at two different points.
• If Δ = 0: The equation has exactly one real solution (a repeated root). The parabola touches the x-axis at only one point—its vertex.
• If Δ < 0: The equation has no real solutions (two complex conjugate solutions). The parabola does not intersect the x-axis.

Applying the Discriminant to Our Problem

Now, let's focus on the specific equation given: x² + ax + a = 0. Comparing this to the standard form ax² + bx + c = 0, we can identify the coefficients:

• a = 1
• b = a
• c = a

Substituting these values into the discriminant formula, we get:

Δ = a² - 4(1)(a) = a² - 4a

For the equation to have exactly one real solution, the discriminant must equal zero:

a² - 4a = 0

Solving for 'a'

This equation is a simple quadratic equation in 'a'. We can solve it by factoring:

a(a - 4) = 0

This equation is satisfied if either:

• a = 0
• a - 4 = 0 => a = 4

So, the equation x² + ax + a = 0 has exactly one real solution when a = 0 or a = 4.

Examining the Solutions: a = 0 and a = 4

Let's investigate each solution separately:

Case 1: a = 0

If a = 0, the equation becomes:

x² = 0

This equation has only one solution: x = 0.

Case 2: a = 4

If a = 4, the equation becomes:

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x² + 4x + 4 = 0

This is a perfect square trinomial, which can be factored as:

(x + 2)² = 0

This equation also has only one solution: x = -2.

Graphical Interpretation

Visualizing these scenarios graphically provides further insight. The equation represents a family of parabolas, each determined by a different value of 'a'. And when a = 0, the parabola is simply y = x², which touches the x-axis at only one point (0,0). When a = 4, the parabola y = x² + 4x + 4 is a downward-opening parabola whose vertex lies on the x-axis at the point (-2, 0). In both cases, the parabola is tangent to the x-axis, indicating a single real root.

The Importance of Considering All Possible Solutions

It's vital to note that both values of 'a' (0 and 4) are valid solutions to the problem. This case highlights the importance of systematically checking for all possibilities. Which means often, when solving equations, students focus on finding a solution, instead of finding all solutions. Neglecting a = 0 would lead to an incomplete answer.

Further Exploration: Analyzing the Behavior of the Parabolas

Let's delve deeper into the behavior of the parabolas for different values of 'a'. The vertex of the parabola y = x² + ax + a is located at x = -a/2. Substituting this into the equation, we find the y-coordinate of the vertex:

y = (-a/2)² + a(-a/2) + a = a²/4 - a²/2 + a = a - a²/4 = a(1 - a/4)

The parabola intersects the x-axis when y = 0, which leads us back to our original equation a² - 4a = 0. This analysis reinforces the fact that the vertex lies on the x-axis only when a = 0 or a = 4.

For values of 'a' other than 0 and 4, the discriminant will be non-zero, resulting in either two distinct real roots or two complex roots. The graphical representation would show parabolas that either intersect the x-axis at two points or do not intersect it at all.

Frequently Asked Questions (FAQ)

Q: Can we solve this problem using the quadratic formula?

A: Yes, absolutely! The quadratic formula provides a direct way to find the roots of a quadratic equation. For x² + ax + a = 0, the quadratic formula gives:

x = [-a ± √(a² - 4a)] / 2

For there to be only one real solution, the discriminant (a² - 4a) must be zero, leading us to the same solutions: a = 0 and a = 4.

Q: What if the equation were slightly different, for example, x² + ax - a = 0?

A: The approach would be similar. Day to day, we would substitute the coefficients into the discriminant, set it equal to zero, and solve for 'a'. The resulting values of 'a' would yield different parabolas with only one x-intercept.

Q: Is there a geometrical interpretation beyond the parabola's tangency to the x-axis?

A: Yes, consider the parabola’s vertex. So naturally, the condition for exactly one solution implies that the vertex of the parabola lies on the x-axis. This provides a valuable geometric visualization.

Conclusion

The problem of determining the value of 'a' for which x² + ax + a = 0 has exactly one real solution elegantly showcases the interplay between algebraic manipulation and geometric interpretation. By utilizing the discriminant and understanding the nature of quadratic equations, we've systematically found that a = 0 and a = 4 are the solutions. Still, this seemingly simple problem encourages a deeper understanding of quadratic equations, their solutions, and the graphical representation of parabolas, enriching one's mathematical intuition and problem-solving skills. Beyond that, it highlights the importance of thoroughly exploring all possible solutions and using multiple approaches to verify findings. Remember to always check your work and consider alternative methods to build a strong foundation in mathematics.