Location of Roots: Quadratic Equation, Theorem, Formula, Questions

Understanding the location of roots of quadratic equations is a vital concept in algebra that plays a significant role in solving mathematical problems across physics, engineering, and economics. In mathematics, this topic helps identify whether the roots are positive, negative, real, or lie on opposite sides of the origin. For students preparing for competitive exams or looking to strengthen their algebra basics, mastering the location of roots in quadratic equations is essential. In this article, we explain the concept with clear explanations, examples, and key conditions to help you grasp it easily.

This Story also Contains

1. Introduction to Location of Roots
2. Standard Form of a Quadratic Equation
3. Different Cases of Roots of a Quadratic Equation
4. Sum and Product of Roots of a Quadratic Equation
5. Graphical Interpretation of Location of Roots
6. Practice Questions on location of roots

Introduction to Location of Roots

The location of roots helps determine where the solutions (roots) of a quadratic equation lie on the number line—whether they are positive, negative, real, complex, or on opposite sides of a point (like the origin or a specific number). Instead of finding the exact values of the roots, the goal is to infer their signs and relative positions using logical conditions and basic algebraic tools.

What is the Location of Roots?

Given a quadratic equation:

$ax^2 + bx + c = 0$

Let the roots be $x_1$ and $x_2$. The location of roots refers to whether:

• Both roots are positive
• Both roots are negative
• Roots lie on opposite sides of zero or another number
• Roots are real and distinct, equal, or complex

Key formulas to analyze root location:

• Discriminant: $D = b^2 - 4ac$
• Sum of roots: $x_1 + x_2 = -\frac{b}{a}$
• Product of roots: $x_1 \cdot x_2 = \frac{c}{a}$

Basic conditions:

• Roots are real and distinct if $D > 0$
• Roots are real and equal if $D = 0$
• Roots are complex (non-real) if $D < 0$
• Both roots are positive if $x_1 + x_2 > 0$ and $x_1 \cdot x_2 > 0$
• Both roots are negative if $x_1 + x_2 < 0$ and $x_1 \cdot x_2 > 0$
• Roots lie on opposite sides of 0 if $x_1 \cdot x_2 < 0$

Standard Form of a Quadratic Equation

Quadratic equations are algebraic expressions of degree 2. Understanding their structure and solutions is key before analyzing root locations.

The standard form of a quadratic equation is: $ax^2 + bx + c = 0$

Where:

• $a$, $b$, and $c$ are real constants
• $a \ne 0$ (since the equation must be quadratic)

Meaning of Roots or Zeros of a Quadratic Equation

The roots (or zeros) of a quadratic equation are the values of $x$ that satisfy:

$ax^2 + bx + c = 0$

These are the x-intercepts of the parabola represented by the function $f(x) = ax^2 + bx + c$.

Roots are calculated using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Real vs Complex Roots

The discriminant $D = b^2 - 4ac$ determines the nature of the roots:

• If $D > 0$, roots are real and distinct
• If $D = 0$, roots are real and equal
• If $D < 0$, roots are complex conjugates

This classification is crucial for analyzing where and how the roots appear on the number line or complex plane.

Conditions for Location of Roots in Quadratic Equations

To determine the sign or side of the roots, certain algebraic tests and relationships involving the coefficients and discriminant are used.

Discriminant and Nature of Roots

As defined:

$D = b^2 - 4ac$

Use it to decide:

• $D > 0$: Two real and distinct roots
• $D = 0$: Real and equal roots
• $D < 0$: No real roots (complex)

Real roots are necessary to discuss their location on the number line.

Sign of Roots Based on Coefficients

Using formulas:

• Sum of roots: $x_1 + x_2 = -\frac{b}{a}$
• Product of roots: $x_1 \cdot x_2 = \frac{c}{a}$

Conditions:

• If $x_1 + x_2 > 0$ and $x_1 \cdot x_2 > 0$ ⇒ both roots are positive
• If $x_1 + x_2 < 0$ and $x_1 \cdot x_2 > 0$ ⇒ both roots are negative
• If $x_1 \cdot x_2 < 0$ ⇒ roots lie on opposite sides of zero

Location of Roots on the Number Line

To determine where the roots lie:

• Check sign of $f(x)$ at a point (like $f(0) = c$)
• Analyze the sign of the product $x_1 \cdot x_2$
• Graphical approach: roots are x-intercepts of the parabola

Quick checks:

• $f(0) = c > 0$ and $x_1 \cdot x_2 < 0$ : roots are on opposite sides of 0
• $x_1 + x_2$ and $x_1 \cdot x_2$ both positive : roots lie on the positive side
• Both positive or both negative depends on sign consistency of the function

Importance in Algebra and Real-World Applications

The concept of location of roots is widely used in:

• Algebra: For solving inequalities and analyzing signs of expressions without solving the equation fully.
• Graphing: Understanding whether the parabola intersects the x-axis and where.
• Inequalities: For checking when $ax^2 + bx + c > 0$ or $< 0$ based on root positions.
• Physics and Engineering: To ensure solutions (e.g., time, speed) are positive and physically meaningful.
• Economics: For identifying profit/loss break-even points based on quadratic cost/revenue functions.

Different Cases of Roots of a Quadratic Equation

The nature and location of the roots of a quadratic equation $ax^2 + bx + c = 0$ depend mainly on the discriminant $D = b^2 - 4ac$. Below are the different cases:

Case 1: $D > 0$ – Real and Distinct Roots

• Roots are real and unequal.
• If $\frac{c}{a} > 0$ and $\frac{-b}{a} > 0$, both roots are positive.
• If $\frac{c}{a} > 0$ and $\frac{-b}{a} < 0$, both roots are negative.
• If $\frac{c}{a} < 0$, roots lie on opposite sides of origin.

Case 2: $D = 0$ – Real and Equal Roots

• Roots are real and repeated
• Single root is given by: $x = \frac{-b}{2a}$
• Root is positive if $\frac{-b}{2a} > 0$, and negative if $\frac{-b}{2a} < 0$.

Case 3: $D < 0$ – Imaginary Roots

• 
  
• Roots are non-real (complex conjugate pair).
• 
  
• No real root exists ⇒ no location on the number line.

Case 4: $D$ is a Perfect Square

• Roots are real and rational
• Examples: $D = 4$, $9$, $16$, etc.
• Occurs often in factorable quadratic equations

Case 5: $D$ is Not a Perfect Square

• Roots are real and irrational.
• Cannot be simplified to a rational number.
• Roots still follow the same sign rules using sum and product.

Sum and Product of Roots of a Quadratic Equation

Let $\alpha$ and $\beta$ be the roots of the quadratic equation:

$ax^2 + bx + c = 0$

Then:

Sum of Roots Formula

$\alpha + \beta = \frac{-b + \sqrt{D}}{2a} + \frac{-b - \sqrt{D}}{2a}$

$= \left( \frac{-b}{2a} + \frac{\sqrt{D}}{2a} \right) + \left( \frac{-b}{2a} - \frac{\sqrt{D}}{2a} \right)$

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

So, sum of roots:

$\alpha + \beta = \frac{-b}{a}$

Product of Roots Formula

$\alpha \cdot \beta = \left( \frac{-b + \sqrt{D}}{2a} \right) \cdot \left( \frac{-b - \sqrt{D}}{2a} \right)$

$= \frac{(-b)^2 - (\sqrt{D})^2}{(2a)^2}= \frac{b^2 - D}{4a^2}$

$= \frac{b^2 - (b^2 - 4ac)}{4a^2}$

$ = \frac{4ac}{4a^2} = \frac{c}{a}$

So, product of roots:

$\alpha \cdot \beta = \frac{c}{a}$

Forming a Quadratic Equation from Roots

Given $\alpha + \beta$ and $\alpha \cdot \beta$, the quadratic equation is:

$x^2 - (\alpha + \beta)x + (\alpha \cdot \beta) = 0$

Substitute values:

$x^2 - \frac{b}{a}x + \frac{c}{a} = 0$

Multiply entire equation by $a$:

$ax^2 - bx + c = 0$

This is the original quadratic equation reconstructed using sum and product of roots.

NCERT Useful Resources

Explore essential NCERT study materials for Complex Numbers and Quadratic Equations, including detailed solutions, concise revision notes, and curated exemplar problems. These resources are tailored to help you strengthen your conceptual understanding and prepare effectively for board and competitive exams.

NCERT Solutions for Chapter 4 Complex Numbers and Quadratic Equations

NCERT Notes for Chapter 4 Complex Numbers and Quadratic Equations

NCERT Exemplar for Chapter 4 Complex Numbers and Quadratic Equations

Graphical Interpretation of Location of Roots

Now, the question is How to find the location of roots? The graphical interpretation of the location of roots helps visualize where the roots of a quadratic equation lie on the x-axis. For any quadratic equation of the form $f(x) = ax^2 + bx + c$, the graph is a parabola, and the x-intercepts represent the real roots.

Whether the parabola cuts the x-axis at two points, touches it once, or doesn’t intersect at all depends on the discriminant. This visual approach makes it easier to understand the nature and position of the roots—positive, negative, repeated, or imaginary.

Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}{ }^2+\mathrm{bx}+\mathrm{c}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are real numbers and ' a ' is non-zero number. Let $x_1$ and $x_2$ be the roots of the equation, and let k be a real number. Then:

1. If both roots of $\mathrm{f}(\mathrm{x})$ are less than k then

i) $\mathrm{D} \geq 0$ (as the real roots may be distinct or equal)
ii) $af( k)>0$ (In both the cases $af(\mathrm{k})$ is positive, as in second case if a $<0$ then $\mathrm{f}(\mathrm{k})<0$, so multiplying two $-ve$ values will give us a positive value)
iii) $k>\frac{-b}{2 a}$ since $\frac{-b}{a}$ will lie between $x_1$ and $x_2$, and $x_1$, $x_2$ are less than k so $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will be less than k .

2. If both roots of $f(x)$ are greater than $k$

i) $\mathrm{D} \geq 0$ (as the real roots may be distinct or equal)
ii) $\mathrm{af}(\mathrm{k})>0$ (In both the cases af( k$)$ is positive, as in second case if a $<0$ then $\mathrm{f}(\mathrm{k})<0$, so multiplying two -ve values will give us a positive value)
iii) $k<\frac{-\mathrm{b}}{2 \mathrm{a}}$ since $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will lie between $x_1$ and $x_2$, and $x_1$, $x_2$ are greater than k so $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will be greater than k .

Condition for number $k$

Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}$ where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are real numbers and ' a ' is non-zero number. Let $x_1$ and $x_2$ be the real roots of the function. And let k is any real number. Then:

3. If $k$ lies between the root $x_1$ and $x_2$

$\mathrm{af}(\mathrm{k})<0$.

As if $a<0$ then $f(k)>0$. So multiplying one $-ve$ and one $+ve$ value will give us negative value)

Condition for number $k_1$ and $k_2$

Let $f(x)=a x^2+b x+c$ where a,b,c are real numbers and ‘a’ is non-zero number. Let $x_1$ and $x_2$ be the real roots of the function. And let $k_1, k_2$ be any two real numbers. Then:

4. If exactly one root of $f(x)$lies in between the number $k_1, k_2$

$f\left(k_1\right) f\left(k_2\right)<0$ as for one value of k, we will have +ve value of$f(x)$ and for other values of $k$, we will have $-ve$ value of $f(x)$ (here $x_1$ < $x_2$)

Condition on number $k_1, k_2$

Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}{ }^2+\mathrm{bx}+\mathrm{c}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are real numbers and '$a$ ' is non-zero number. Let $x_1$ and $x_2$ be the roots of the function. And let $k 1, k_2$ be any two real numbers. Then

5. If both roots lie between $k_1, k_2$

i) $\mathrm{D} \geq 0$ (as the real roots may be distinct or equal)
ii) $k_1<\frac{-b}{2 a}<k_2$, where $a \leq 3$ and $k_1<k_2$

6. If $k_1, k_2$ lies between the roots

$\mathrm{af}\left(\mathrm{k}_1\right)<0$ and $\mathrm{af}\left(\mathrm{k}_2\right)<0$

Solved Examples Based on Location of Roots

Example 1: If the roots of the quadratic equation $x^2 - 3x + k = 0$ lie on opposite sides of zero, find the range of $k$.

Solution: For roots to lie on opposite sides of zero:

$\Rightarrow x_1 \cdot x_2 < 0$

Here, $a = 1$, $b = -3$, $c = k$

Use the product of roots formula: $x_1 \cdot x_2 = \dfrac{c}{a} = \dfrac{k}{1} = k$

So, $k < 0$

Example 2: Find the value of $k$ for which both roots of the equation $x^2-4 k x-4-k+9 k^2=0$ are positive

1) $k \in\left(-\frac{4}{5}, 1\right)$

2) $k \in \phi$

3) $k>\frac{5}{2}$

4) $k \in\left(-\infty, \frac{7-\sqrt{135}}{6}\right) \cup\left(\frac{7+\sqrt{135}}{6}, \infty\right)$

Solution

i) $D \geq 0$

$(-4 k)^2-4\left(-4-k+9 k^2\right) \geq 0$

$16 k^2+16+4 k-36 k^2 \geq 0$

$5 k^2-k-4 \leq 0$

$k \in\left[-\frac{4}{5}, 1\right]$

ii) $a.f(0)>0$

$0-4 k(0)-4-k+9 k^2>0$

From (i), (ii) and (iii)

$k \in \phi$ or for no values of k sum of the roots is greater than 5.

Example 3: The set of all real values of $\lambda$ for which the quadratic equations,

$\left(\lambda^2+1\right) x^2-4 \lambda x+2=0$ always have exactly one root in the interval $(0,1)$ is :

1) $(0,2)$
2) $(2,4)$
3) $(1,3)$
4) $(-3,-1)$

Solution:

If exactly one root in $(0, 1)$ then

$\Rightarrow \mathrm{f}(0) \cdot \mathrm{f}(1)<0$

$\Rightarrow 2\left(\lambda^2-4 \lambda+3\right)<0$

$\Rightarrow 1<\lambda<3$

Now for $\lambda=1$

$2 x^2-4 x+2=0$

$(x-1)^2=0 $

$x=1.1$

So both roots doesn’t lie between (0, 1)

$\therefore \lambda \neq 1$

$ \text { Again for } \lambda=3$

$10 x^2-12 x+2=0 $

$\Rightarrow x=1, \frac{1}{5}$

So if one root is $1$ then second root lie between $(0, 1)$ so $\lambda=3$

$\therefore \lambda \in(1,3)$

Example 3: If both the roots of the quadratic equation $x^2-2 p x+p^2+p-5=0$ are less than $3$, then the largest integral value of $p$ is________.

1)$ 0$

2) $1$

3) $2$

4) $3$

Solution:
$-\frac{b}{2 a}<3$
$\Rightarrow \frac{2 p}{2}<3$
$\Rightarrow p<3$
Also D $>0$
$1 p^2-4\left(p^2+p-5\right) \geq 0$

Also, $5-6 p+p^2+p-5>0$
$\Rightarrow p^2-5 p+4>0$
$\Rightarrow(p-1)(p-4)>0$
From (i), (ii),(iii)

$p \in(-\infty, 1)=0$
Hence, the answer is the option 1.

Example 4: For what values of $k$ are both roots of $x^2 - (k + 1)x + k = 0$ positive?

Solution:Let $a = 1$, $b = -(k + 1)$, $c = k$

To have both roots positive, the following conditions must be satisfied:

Discriminant > 0:

$D = b^2 - 4ac = (k + 1)^2 - 4k = k^2 + 2k + 1 - 4k = k^2 - 2k + 1 = (k - 1)^2$

Since $(k - 1)^2 > 0$ for $k \ne 1$,
$\Rightarrow k \ne 1$

Sum of roots > 0:

$x_1 + x_2 = \dfrac{-b}{a} = \dfrac{k + 1}{1} = k + 1 > 0 \Rightarrow k > -1$

Product of roots > 0:

$x_1 \cdot x_2 = \dfrac{c}{a} = \dfrac{k}{1} = k > 0$

So combining: $k > 0$ (stronger than $k > -1$)
$k \ne 1$

Example 5: Find the values of $m$ such that both roots of $x^2 + mx + 3 = 0$ are less than $2$.

Given:
Let $f(x) = x^2 + mx + 3$

We want both roots less than 2. Conditions:

1. $D \geq 0$:
   $D = m^2 - 4ac = m^2 - 12 \geq 0 \Rightarrow m \leq -2\sqrt{3}$ or $m \geq 2\sqrt{3}$
2. $f(2) > 0$:
   $f(2) = 4 + 2m + 3 = 2m + 7 > 0 \Rightarrow m > -\dfrac{7}{2}$
3. Vertex < 2:
   $\dfrac{-m}{2} < 2 \Rightarrow m > -4$

Combine all:

• $m \geq 2\sqrt{3}$ satisfies all
• $m \leq -2\sqrt{3}$ is valid only if $m > -\dfrac{7}{2}$ and $m > -4$

So final answer:

$m \leq -2\sqrt{3} \quad \text{or} \quad m \geq 2\sqrt{3}$

Practice Questions on location of roots

To help you strengthen your understanding of the topic, we’ve included a few practice questions on location of roots of quadratic equations. These will test your grasp of root positions, signs, and related conditions using standard formulas.

To practice location of roots based questions, click here.

You can practice the next topics of Complex Numbers and Quadratic Equations below:

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