Nature of Roots and the Discriminant for CBSE Class 10 Mathematics

In algebra, solving an equation gives us its exact numerical roots. However, in many engineering, physical, and board examination contexts, we do not need the exact numerical values of the roots; we only need to know their nature—whether the roots are real and distinct, real and equal, or non-existent in the real number system.

The quantity that reveals this information without requiring us to solve the equation is the discriminant. In CBSE Class 10 Mathematics, questions on the discriminant—especially finding an unknown parameter $k$ for which a quadratic equation has equal roots—are among the most reliable, guaranteed questions in the board exam.

What You Will Learn

• Definition and significance of the Discriminant ($D = b^2 - 4ac$)
• The three distinct cases governing the nature of roots ($D > 0, D = 0, D < 0$)
• Geometric connection: how the discriminant predicts the parabola's $x$-intercepts
• The condition for real and equal roots ($D = 0$)
• Step-by-step methods to solve board exam questions finding unknown parameter $k$
• Common traps involving parameter constraints and division by zero

1. What is the Discriminant?

For a standard quadratic equation $ax^2 + bx + c = 0$ (where $a \ne 0$), the quadratic formula expresses the roots as: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The nature of these roots is entirely determined by the expression lying under the square root radical sign, $b^2 - 4ac$. Because it discriminates between different types of roots, it is called the discriminant, denoted by the letter $D$:

$D = b^2 - 4ac$

The discriminant tells us about the roots before we solve the equation.

2. The Three Cases for the Nature of Roots

                           The Discriminant: D = b² - 4ac
                                         |
       +---------------------------------+---------------------------------+
       |                                 |                                 |
     D > 0                             D = 0                             D < 0
Two Distinct Real Roots            Two Equal Real Roots               No Real Roots
(Parabola crosses x-axis twice)   (Parabola touches x-axis once)   (Parabola never touches x-axis)

Case 1: When $D > 0$ (Two Distinct Real Roots)

• Since $D$ is strictly positive, $\sqrt{D}$ is a real number greater than $0$.
• The formula yields two different values: $x_1 = \frac{-b + \sqrt{D}}{2a}, \quad x_2 = \frac{-b - \sqrt{D}}{2a}$
• Conclusion: The equation has two distinct real roots.
• Graphical Meaning: The parabolic graph of $y = ax^2 + bx + c$ intersects the $x$-axis at two distinct points.

Case 2: When $D = 0$ (Two Equal / Coincident Real Roots)

• Since $D = 0$, $\sqrt{D} = 0$.
• Both roots collapse to the single value: $x_1 = x_2 = -\frac{b}{2a}$
• Conclusion: The equation has two equal (coincident) real roots.
• Graphical Meaning: The parabola touches the $x$-axis at exactly one point (the vertex of the parabola lies on the $x$-axis).

Important: <u>If a question states that an equation has "real roots", this means $D \ge 0$ (combining both Case 1 and Case 2)! If it states "equal roots", it means strictly $D = 0$.</u>

Case 3: When $D < 0$ (No Real Roots)

• Since $D$ is negative, $\sqrt{D}$ is the square root of a negative number, which is not a real number.
• Conclusion: The equation has no real roots (its roots are complex/imaginary numbers).
• Graphical Meaning: The parabola lies entirely above the $x$-axis ($a > 0$) or entirely below the $x$-axis ($a < 0$), never touching or crossing it.

3. High-Yield Board Exam Problems: Finding Unknown Parameter $k$

Questions requiring students to find $k$ for equal roots appear almost every year in CBSE board exams.

Solved Example 1: Standard Parameter Equation

Problem: Find the values of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two equal roots.

Solution:

1. Identify coefficients from $ax^2 + bx + c = 0$: $a = 2, \quad b = k, \quad c = 3$
2. Compute the discriminant $D$: $D = b^2 - 4ac = (k)^2 - 4(2)(3) = k^2 - 24$
3. For the equation to have two equal roots, the discriminant must be zero ($D = 0$): $k^2 - 24 = 0$ $k^2 = 24$ $k = \pm \sqrt{24} = \pm \sqrt{4 \times 6} = \pm 2\sqrt{6}$
4. Therefore, <u>$k = 2\sqrt{6}$ or $k = -2\sqrt{6}$</u>.

Solved Example 2: The Leading Coefficient Trap (CBSE Classic)

Problem: Find the value of $k$ for which the equation $kx(x - 2) + 6 = 0$ has two equal roots.

Solution:

1. First expand and write in standard form: $kx^2 - 2kx + 6 = 0$
2. Identify coefficients: $a = k, \quad b = -2k, \quad c = 6$
3. Compute the discriminant: $D = b^2 - 4ac = (-2k)^2 - 4(k)(6) = 4k^2 - 24k$
4. For equal roots, set $D = 0$: $4k^2 - 24k = 0$ $4k(k - 6) = 0$
5. This gives two possible values: $k = 0 \quad \text{or} \quad k = 6$
6. Critical Check: Look back at the original equation $kx^2 - 2kx + 6 = 0$. If $k = 0$, the coefficient of $x^2$ becomes zero ($a = 0$), and the equation reduces to $6 = 0$, which is impossible and not a quadratic equation! Therefore, <u>$k = 0$ must be rejected</u>.
7. Hence, the only valid solution is $k = 6$.

Exam Tip: Whenever the parameter $k$ appears in the coefficient of $x^2$, always check whether $k = 0$ eliminates the quadratic term. If it does, you must explicitly reject it to earn the final answer mark!

4. Master Summary Table

Common Mistake: Writing $\pm$ only when simplifying numbers, but forgetting it for parameters. In $k^2 = 16$, $k = \pm 4$ (both $+4$ and $-4$). Writing only $k = 4$ loses half a mark!