Quadratic Equations with Parameters

Definition

1.Quadratic equations are usually introduced in their classical form, where all coefficients are fixed real numbers. The standard expression serves as the basic model for every second-degree equation:

$$ax^{2} + bx + c = 0$$

2.The key ideas behind quadratic equations are closely linked to the geometry of the parabola and to the behaviour of the discriminant. These elements determine how the graph bends, where its vertex lies, and whether the equation has two real solutions, a repeated solution, or a pair of complex roots.

3.In many situations, however, the coefficients are not constant but depend on an external quantity treated as a parameter. The numbers $a$, $b$, and $c$ then become functions of a real parameter $k$, and the equation takes the more general form:

$$a(k)x^{2} + b(k)x + c(k) = 0, \qquad a(k) \neq 0$$

4.The condition $a(k) \neq 0$ prevents the equation from collapsing into a linear equation with parameters as the parameter varies, ensuring that it retains the structure of a quadratic for all admissible values of $k$.

5.

This page focuses on the quadratic case. For a broader discussion of parametric equations across different degrees, see equations with parameters.

The role of the discriminant

6.The discriminant describes how the solutions of a quadratic equation behave. When the coefficients depend on a parameter, this idea is unchanged, but the discriminant becomes a function of that parameter. Starting from the coefficients $a(k)$, $b(k)$, and $c(k)$, we obtain:

$$\Delta(k) = b(k)^{2} - 4a(k)c(k)$$

7.In this form the discriminant gives a picture of what happens to the roots for each value of $k$. Its sign determines the nature of the solutions:

Example 1

9.Consider the quadratic equation:

$$x^{2} + kx + 1 = 0$$

10.This example is useful because each coefficient depends on the parameter $k$ in a simple way:

$$a(k) = 1 \qquad b(k) = k \qquad c(k) = 1$$

11.In the standard case the discriminant is $\Delta = b^{2} - 4ac$, and its sign determines the nature of the solutions. Substituting $a(k)$, $b(k)$, and $c(k)$ into the same formula gives:

$$\Delta(k) = b(k)^{2} - 4a(k)c(k)$$

12.which simplifies to:

$$\Delta(k) = k^{2} - 4$$

13.To determine when real solutions exist, we study the quadratic inequality:

$$k^{2} - 4 \ge 0$$

14.Solving it yields two separate intervals:

$$k \le -2 \quad\text{or}\quad k \ge 2$$

15.For these values the discriminant is positive and the quadratic has two distinct real roots. At the boundary points $k = \pm 2$ the discriminant becomes zero and the equation has a single repeated solution. When:

$$-2 < k < 2$$

16.the discriminant is negative and the equation has no real roots, producing instead a pair of complex conjugate solutions. The following table summarises the sign analysis of $k - 2$, $k + 2$, and their product across the three intervals determined by the critical points $k = -2$ and $k = 2$:

18.When the roots are real, an explicit expression for them is useful. Applying the quadratic formula to this family gives the two real solutions in closed form:

$$x = \frac{-k \pm \sqrt{k^{2} - 4}}{2}$$

19.This expression shows how each root depends on the parameter $k$ and makes it possible to study how the solutions move along the real line as $k$ varies. From the analysis above, the outcomes are:

Geometric interpretation of parameter changes

21.Continuing with the equation of Example 1, it is helpful to look at how the graph of the corresponding parabola changes as the parameter $k$ varies. To each value of $k$ we associate the parabola:

$$y = x^{2} + kx + 1$$

22.so analysing the roots is equivalent to studying where this curve meets the $x$-axis. As the parameter moves, the entire family of parabolas shifts smoothly. The behaviour can be described in three regimes.

23.When $|k|$ is very large ($k \to +\infty$ or $k \to -\infty$), the linear term $kx$ dominates and the two roots separate widely. A simple asymptotic argument shows that one root tends to $0$ while the other behaves approximately like $-k$. Geometrically, the intersection points move far apart: one stays close to the origin, while the other drifts along the negative axis, and the parabola appears increasingly tilted.

24.At the critical values $k = \pm 2$, the discriminant becomes zero. This is the moment when the parabola touches the $x$-axis at exactly one point: the graph has a single point of tangency, the equation has one repeated root, and the vertex lies on the axis.

25.Within the interval $-2 < k < 2$, the discriminant is negative, so the equation has no real solutions. Geometrically, the parabola does not meet the $x$-axis at all. Since the leading coefficient is positive, the parabola opens upward, and in this region the vertex lies strictly above the $x$-axis. The entire curve stays on one side of the axis, reflecting that the solutions move into the complex plane.

Example 2

26.Consider a more elaborate parameter-dependent quadratic equation:

$$(a + 2)x^{2} + (3a - 1)x + (a - 4) = 0, \qquad a \neq -2$$

27.where the condition $a \neq -2$ ensures that the coefficient of $x^{2}$ does not vanish, so the expression remains a genuine quadratic for every admissible value of the parameter. The goal is to determine for which values of $a$ the equation has two distinct real solutions. As usual, this depends on the sign of the discriminant. To avoid confusion with the parameter $a$, we denote the coefficients of the present quadratic by $A$, $B$, $C$, so the general formula reads:

$$\Delta = B^{2} - 4AC$$

28.Substituting the coefficients of the present equation gives:

$$\Delta(a) = (3a - 1)^{2} - 4(a + 2)(a - 4)$$

29.Expanding the two parts, we first compute the square:

$$(3a - 1)^{2} = 9a^{2} - 6a + 1$$

30.and then the product of the quadratic and constant coefficients:

$$(a + 2)(a - 4) = a^{2} - 2a - 8$$

31.Inserting these expressions into the discriminant and collecting like terms gives the simplified form:

$$\Delta(a) = 5a^{2} + 2a + 33$$

32.This is a quadratic function of $a$ with positive leading coefficient. To determine whether it can ever be zero or negative, we examine its discriminant:

$$2^{2} - 4 \cdot 5 \cdot 33 = 4 - 660 < 0$$

33.Since this value is negative, the parabola associated with $5a^{2} + 2a + 33$ never meets the horizontal axis: it remains strictly positive for all real values of $a$. Therefore:

$$\Delta(a) > 0 \quad \forall\, a \in \mathbb{R}$$

34.The equation has two distinct real solutions for every real value of the parameter $a$, except $a = -2$, which must be excluded because it would make the coefficient of $x^{2}$ vanish and reduce the expression to a first-degree equation.

Example 3

35.Consider the parameter-dependent quadratic equation:

$$x^{2} - (a - 4)x + (a^{2} - a) = 0$$

36.where $a$ is a real parameter. We want the values of $a$ for which the product of the two solutions is strictly less than $2$. For a general monic quadratic equation:

$$x^{2} + bx + c = 0$$

37.the product of the solutions satisfies $x_{1}x_{2} = c$.

38.

This identity follows from the factorised form of a monic quadratic, where $x^{2} + bx + c = (x - x_{1})(x - x_{2})$. Expanding the product, the constant term appears as $x_{1}x_{2}$, which is why the product of the solutions equals $c$.

39.Here the constant term is $c = a^{2} - a$, and requiring the product of the two solutions to be less than $2$ gives the inequality:

$$a^{2} - a < 2$$

40.Bringing all terms to the left gives:

$$a^{2} - a - 2 < 0$$

41.The quadratic polynomial on the left factors as:

$$a^{2} - a - 2 = (a - 2)(a + 1)$$

42.The critical points are therefore $a = -1$ and $a = 2$. To determine where the expression is negative, we analyse the signs of the two factors:

44.The sign chart shows that $(a - 2)(a + 1) < 0$ when the parameter lies between the two roots:

$$-1 < a < 2$$

45.The product of the solutions satisfies $x_{1}x_{2} < 2$ if and only if $-1 < a < 2$.