Nature of roots — discriminant based classification with graphical interpretation

Question

How do we determine the nature of roots of a quadratic equation using the discriminant, and what does each case look like graphically?

Solution — Step by Step

For the quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$ ):

D = b^2 - 4ac

The discriminant $D$ completely determines the nature of the roots. We do not need to actually solve the equation.

Case 1: $D > 0$ — Two distinct real roots

x = \frac{-b \pm \sqrt{D}}{2a}

The two roots are different real numbers. Graphically, the parabola crosses the x-axis at two distinct points.

Subcase: If $D$ is a perfect square AND $a$ , $b$ , $c$ are rational, then the roots are rational. Otherwise, they are irrational (involving $\sqrt{D}$ ).

Case 2: $D = 0$ — Two equal real roots (repeated root)

x = \frac{-b}{2a}

Both roots are the same. Graphically, the parabola just touches the x-axis at one point (the vertex).

Case 3: $D < 0$ — No real roots

The square root of a negative number is not real. The equation has no real solutions. Graphically, the parabola does not intersect the x-axis at all — it floats entirely above (if $a > 0$ ) or below (if $a < 0$ ).

Find the nature of roots of $2x^2 - 5x + 3 = 0$ .

$a = 2$ , $b = -5$ , $c = 3$

D = (-5)^2 - 4(2)(3) = 25 - 24 = 1

Since $D = 1 > 0$ and $D$ is a perfect square with rational coefficients, the roots are two distinct rational roots.

Actual roots: $x = \frac{5 \pm 1}{4}$ , giving $x = \frac{3}{2}$ and $x = 1$ .

Example: For what values of $k$ does $x^2 + kx + 9 = 0$ have equal roots?

Equal roots means $D = 0$ :

k^2 - 4(1)(9) = 0

k^2 = 36

k = \pm 6

This type of “find $k$ ” question is extremely common in CBSE 10th and JEE Main.

flowchart TD
    A["ax2 + bx + c = 0"] --> B["Calculate D = b2 minus 4ac"]
    B --> C{"Value of D?"}
    C -->|"D greater than 0"| D["Two distinct real roots"]
    C -->|"D = 0"| E["Two equal roots: x = minus b over 2a"]
    C -->|"D less than 0"| F["No real roots"]
    D --> G{"Is D a perfect square?"}
    G -->|"Yes"| H["Rational roots"]
    G -->|"No"| I["Irrational roots"]

Why This Works

The quadratic formula has $\sqrt{D}$ in it. When $D > 0$ , the square root is a positive real number, giving two different values ( $-b + \sqrt{D}$ and $-b - \sqrt{D}$ ). When $D = 0$ , both values become $-b/(2a)$ . When $D < 0$ , we are taking the square root of a negative number, which has no real value — hence no real roots.

The discriminant is like a traffic signal for quadratic equations: it tells you what to expect without solving.

Alternative Method

For JEE-level problems, combine the discriminant with Vieta’s formulas (sum and product of roots):

\alpha + \beta = -\frac{b}{a}, \qquad \alpha \beta = \frac{c}{a}

If the question asks about the nature of roots in terms of positive/negative or about the signs of roots, Vieta’s formulas are often more useful than the discriminant alone. For example, if $D > 0$ and $c/a > 0$ and $-b/a > 0$ , both roots are positive.

Common Mistake

Students sometimes compute $D = b^2 + 4ac$ (adding instead of subtracting). The discriminant is $b^2 - 4ac$ , always with a minus sign. A wrong sign flips the conclusion entirely — a negative discriminant becomes positive, and you would incorrectly claim real roots exist when they do not. Write the formula carefully every time, especially under exam pressure.