For a quadratic equation, ax 2+ bx + c = 0, if b 2– 4ac = 0, then the roots are,
Real and equal
Understanding Quadratic Equation Roots
A standard quadratic equation is written in the form:
$\qquad ax^2 + bx + c = 0$
where $a$, $b$, and $c$ are coefficients and $a \neq 0$. The values of $x$ that satisfy this equation are called its roots.
The Role of the Discriminant
The nature of the roots of a quadratic equation is determined by a value called the discriminant. The discriminant is calculated using the formula:
$\qquad \Delta = b^2 - 4ac$
The value of the discriminant tells us whether the roots are real or imaginary, and whether they are distinct (different) or equal (the same).
Analyzing the Given Condition: $b^2 - 4ac = 0$
The question states that for the quadratic equation $ax^2 + bx + c = 0$, the discriminant is zero:
$\qquad b^2 - 4ac = 0$
Let's look at the quadratic formula, which gives the roots of the equation:
$\qquad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
If we substitute $b^2 - 4ac = 0$ into this formula, we get:
$\qquad x = \frac{-b \pm \sqrt{0}}{2a}$
$\qquad x = \frac{-b \pm 0}{2a}$
This simplifies to:
$\qquad x = \frac{-b}{2a}$
Since the part under the square root is zero, there is no ± variation yielding two different values. Both roots are given by the single expression $\frac{-b}{2a}$.
Therefore, when the discriminant $b^2 - 4ac = 0$, the roots are:
- Real: The value $\frac{-b}{2a}$ is a real number since $b$ and $a$ are real coefficients (as implied by the standard form).
- Equal: Both roots have the same value, $\frac{-b}{2a}$.
Comparing Discriminant Cases
Here's a summary of how the discriminant affects the nature of the roots:
| Discriminant ($\Delta$) | Nature of Roots |
|---|---|
| $\Delta > 0$ ($b^2 - 4ac > 0$) | Real and distinct (unequal) |
| $\Delta = 0$ ($b^2 - 4ac = 0$) | Real and equal |
| $\Delta < 0$ ($b^2 - 4ac < 0$) | Imaginary (complex conjugates) and distinct |
Based on our analysis and the table, when $b^2 - 4ac = 0$, the roots of the quadratic equation are real and equal.
Revision Table: Quadratic Roots and Discriminant
| Concept | Formula/Condition | Outcome |
|---|---|---|
| Quadratic Equation | $ax^2 + bx + c = 0$ | Defines a second-degree polynomial equation |
| Discriminant | $\Delta = b^2 - 4ac$ | Determines the nature of roots |
| Roots Formula | $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ | Provides the values of the roots |
| Condition for Real and Equal Roots | $\Delta = 0 \implies b^2 - 4ac = 0$ | Roots are real numbers and have the same value |
Additional Information: Geometric Interpretation
The roots of a quadratic equation $ax^2 + bx + c = 0$ represent the x-intercepts of the parabola $y = ax^2 + bx + c$.
- If $\Delta > 0$, the parabola intersects the x-axis at two distinct points (real and distinct roots).
- If $\Delta = 0$, the parabola touches the x-axis at exactly one point (real and equal roots - the vertex is on the x-axis).
- If $\Delta < 0$, the parabola does not intersect the x-axis (imaginary and distinct roots).
Understanding the discriminant is key to quickly determining the nature of the solutions to any quadratic equation without fully solving it.
Important Questions from Quadratic Equations
- The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation, $10x^2 - 27x + \alpha = 0$ are rational numbers is:
The sum of all real values of x satisfying the equation
\(\rm (x^2 - 5x + 5) ^{x^2 + 4x - 60 }= 1\) is:
The number of integral values of $m$ for which the quadratic expression, $(10m-9)x^2 - 2mx + 1$, where $x \in \mathbb{R}$, is always positive, is
It is given that the equations x 2– y 2= 0 and (x – a) 2+ y 2= 1 have single positive solution. For this, the value of ‘a’ is
If α and β are the roots of the quadratic equation 2x 2+ 6x + k = 0, where k < 0, then what is the maximum value of (α/β + β/α)?
