Quadratic Equations

Definition

A quadratic equation in one variable is a second-degree polynomial equation of the form:

\( ax^2 + bx + c = 0 \)

where a, b, c ∈ ℝ and a ≠ 0. The condition \(a ≠ 0\) ensures the equation remains quadratic (degree = 2).

Understanding the Standard Form

• a → coefficient of \(x^2\) (controls shape of graph)
• b → coefficient of \(x\) (controls axis shift)
• c → constant term (y-intercept)

Examples (Correct and Board-Level)

\[ \begin{aligned} &2x^2 - 3x - 5 = 0 \\ &x^2 = 0 \\ &-3x^2 + 9x + 1 = 0 \\ &x(x-4) + 4 = 0 \Rightarrow x^2 - 4x + 4 = 0 \end{aligned} \]

Not Quadratic Equations

• \(2x^3 + 1 = 0\) → degree 3 (cubic)
• \(\frac{1}{x} + x = 0\) → not a polynomial
• \(ax + b = 0\) → linear equation

Graphical Meaning

The graph of a quadratic equation is a parabola. It can open upwards or downwards depending on coefficient \(a\).

The points where the parabola intersects the x-axis represent the roots of the equation.

Types of Quadratic Equations

• Complete Quadratic Equation: All terms present (e.g., \(x^2 + 5x + 6 = 0\))
• Incomplete Quadratic Equation: Missing one term:
  • \(ax^2 + bx = 0\)
  • \(ax^2 + c = 0\)

Roots of Quadratic Equation

The solutions of a quadratic equation are called roots.

• If graph cuts x-axis at two points → two distinct real roots
• If touches x-axis → equal roots
• If does not intersect → no real roots

Importance for Board Exams

• High-weightage chapter (frequently 6–8 marks)
• Forms base for higher mathematics (Class 11 & 12)
• Used in:
  • Word problems
  • Graph-based questions
  • Case-study questions (CBSE pattern)
• Concept directly linked with:
  • Discriminant
  • Nature of roots
  • Factorisation & formula method

Real-Life Applications

• Projectile motion (physics)
• Profit maximization (economics)
• Area optimization problems
• Engineering curve design

Exam-Oriented Tips

• Always convert equation into standard form first
• Check coefficient signs carefully
• Verify roots by substitution
• Prefer factorization when possible (faster in exams)

Concept and Definition

Any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree 2, is called a quadratic equation.

When all terms of \(p(x)\) are arranged in descending order of powers of \(x\), the equation takes its standard (canonical) form:

\( ax^2 + bx + c = 0,\quad a \ne 0 \)

Structure of Standard Form

• Quadratic term: \(ax^2\) (degree 2)
• Linear term: \(bx\) (degree 1)
• Constant term: \(c\) (degree 0)

The coefficient \(a\) must not be zero; otherwise, the equation reduces to a linear equation.

How to Convert Any Equation into Standard Form

1. Expand brackets (if any)
2. Bring all terms to one side so RHS becomes 0
3. Arrange terms in descending powers of \(x\)
4. Simplify coefficients

Examples (Exam-Oriented)

\[ \begin{aligned} &2x(x+3) = 5 \Rightarrow 2x^2 + 6x - 5 = 0 \\ &x(x-1) + 6 = 0 \Rightarrow x^2 - x + 6 = 0 \\ &(x+2)^2 = 9 \Rightarrow x^2 + 4x - 5 = 0 \\ &\frac{1}{x} + x = 2 \Rightarrow \text{Not quadratic (non-polynomial)} \end{aligned} \]

Visual Interpretation of Standard Form

Every quadratic equation in standard form represents a parabola. The coefficients \(a, b, c\) determine its shape, position, and intercepts.

Key Observations

• If b = 0 → equation is of type \(ax^2 + c = 0\)
• If c = 0 → equation becomes \(x(ax + b) = 0\)
• Standard form is required before applying:
  • Quadratic formula
  • Discriminant analysis
  • Factorisation method

Importance for CBSE Board Exams

• Almost every question begins with conversion to standard form
• Common in case-study and word problems
• Required for applying formulas correctly
• Errors in rearrangement lead to incorrect roots

Common Mistakes to Avoid

• Not shifting all terms to one side
• Wrong sign while transposing terms
• Not simplifying completely
• Treating non-polynomial equations as quadratic

Conceptual Insight

Quadratic equations arise when a problem involves a product of two related quantities, squared relationships, or optimization (maximum/minimum).

In most CBSE problems, the equation is not given directly—you must translate words into algebra, which typically leads to a quadratic equation.

General Formation Pattern

Most real-life problems follow this structure:

• Let unknown quantity = \(x\)
• Express related quantity in terms of \(x\)
• Use given condition → form equation
• Simplify to obtain \(ax^2 + bx + c = 0\)

Common Situations Where Quadratic Equations Appear

• Area and Geometry Problems
  • Rectangle dimensions using area → \(x(x+k)\)
  • Square or path problems involving area difference

  Example: Area of rectangle = 60, length = \(x+5\)
  \(x(x+5)=60 \Rightarrow x^2 + 5x - 60 = 0\)

• Number Problems
  • Two numbers with given sum and product
  • Consecutive integers or digits problems

  Example: Sum = 10, product = 21
  Let numbers be \(x\) and \(10-x\)
  \(x(10-x)=21 \Rightarrow x^2 - 10x + 21 = 0\)

• Motion and Time Problems
  • Vertical motion (projectile)
  • Speed-time-distance relations

  Example: Height equation \(h = ut - \frac{1}{2}gt^2\) → quadratic in \(t\)

• Profit, Revenue and Business Problems
  • Revenue = price × quantity
  • Maximum profit scenarios

  Revenue often becomes:
  \(R = x(100 - x) \Rightarrow R = 100x - x^2\)

Visual Understanding (Area Model)

Many geometry problems directly form expressions like \(x(x+k)\), which expand into quadratic equations.

CBSE Board Strategy

• Always define variable clearly (marks are awarded)
• Convert statement → equation step-by-step
• Check final answer satisfies real-life condition
• Reject negative or invalid values when required

High-Scoring Insight

If a problem involves:

• Product of two expressions → likely quadratic
• Area or geometry → quadratic almost certain
• Maximum/minimum → parabola concept

Common Mistakes

• Wrong variable assumption
• Incorrect translation of words into equation
• Forgetting units or real-world constraints
• Not rejecting extraneous roots

Overview

A quadratic equation can be solved using multiple methods depending on its structure. Choosing the most efficient method is crucial for saving time in board exams.

Method of Factorization

This method is used when the quadratic expression can be expressed as a product of two linear factors.

Core Idea

Convert
\(ax^2 + bx + c = 0\)
into
\((px + q)(rx + s) = 0\)

Steps

• Compute \(a \times c\)
• Find two numbers whose sum = \(b\) and product = \(ac\)
• Split middle term
• Group and factor
• Apply zero-product property

\[ \text{If } pq = 0 \Rightarrow p = 0 \text{ or } q = 0 \]

Example

\[ \begin{aligned} x^2 + 7x + 10 &= 0 \\ x^2 + 5x + 2x + 10 &= 0 \\ x(x+5) + 2(x+5) &= 0 \\ (x+2)(x+5) &= 0 \\ \Rightarrow x = -2,\ -5 \end{aligned} \]

Best used when: Roots are integers or simple rational numbers.

Completing the Square

This method transforms the quadratic equation into a perfect square form.

Core Idea

Convert equation into:
\((x + p)^2 = q\)

Steps

• Divide entire equation by \(a\) (if \(a \ne 1\))
• Move constant term to RHS
• Add \(\left(\frac{b}{2a}\right)^2\) to both sides
• Rewrite LHS as perfect square
• Take square root and solve

Example

\[ \begin{aligned} x^2 + 6x + 5 &= 0 \\ x^2 + 6x &= -5 \\ x^2 + 6x + 9 &= 4 \\ (x+3)^2 &= 4 \\ x+3 &= \pm 2 \\ x &= -1,\ -5 \end{aligned} \]

Best used when: Equation is not easily factorable.

Quadratic Formula

This is the most general method and works for all quadratic equations.

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

Discriminant

\(D = b^2 - 4ac\)

• D > 0: Two distinct real roots
• D = 0: Equal (repeated) roots
• D < 0: No real roots

Best used when: Factorization is difficult or coefficients are large.

Which Method to Choose?

• If expression is simple → use factorization
• If perfect square pattern visible → use completing square
• If coefficients are complex → use quadratic formula

Exam Strategy

• Always write equation in standard form first
• Try factorization before using formula
• Use discriminant to predict nature of roots
• Verify answers by substitution

Common Mistakes

• Incorrect splitting of middle term
• Sign errors in quadratic formula
• Forgetting ± in square root
• Calculation errors in discriminant

Discriminant (D)

For a quadratic equation \(ax^2 + bx + c = 0\), the nature of roots depends on the discriminant:

\(D = b^2 - 4ac\)

Cases Based on Discriminant

• \(D > 0\): Real and Distinct Roots

  Two different real solutions exist:
  \(x_1,\ x_2 = \frac{-b \pm \sqrt{D}}{2a}\)

  Graph cuts x-axis at two distinct points.

• \(D = 0\): Real and Equal Roots

  Only one repeated root exists:
  \(x = \frac{-b}{2a}\)

  Graph touches the x-axis (tangent).

• \(D < 0\): No Real Roots

  Roots are complex (not visible on real number line).

  Graph does not intersect the x-axis.

Graphical Interpretation (Correct Representation)

The number of intersection points with the x-axis directly equals the number of real roots: 2, 1, or 0.

Important Analytical Insights

• If \(D\) is a perfect square → roots are rational
• If \(D\) is not a perfect square → roots are irrational
• If \(b^2 = 4ac\) → equation is a perfect square

CBSE Board Focus

• Very common 2–3 mark direct question
• Used in case-study questions
• Often combined with quadratic formula
• Asked to determine condition for equal roots

Common Mistakes

• Sign error in \(b^2 - 4ac\)
• Forgetting square root while applying formula
• Misinterpreting D < 0 as “no solution” (actually complex roots exist)

Check whether the following are quadratic equations:

• \((x - 2)^2 + 1 = 2x - 3\)
• \((x + 2)^3 = x^3 - 4\)

Concept Used

An equation is quadratic if it can be reduced to the form:
\(ax^2 + bx + c = 0,\ a \ne 0\)

Solution (i)

\[ \begin{aligned} (x - 2)^2 + 1 &= 2x - 3 \\ x^2 - 4x + 4 + 1 &= 2x - 3 \\ x^2 - 4x + 5 - 2x + 3 &= 0 \\ x^2 - 6x + 8 &= 0 \end{aligned} \]

The equation is in the form \(ax^2 + bx + c = 0\). Hence, it is a quadratic equation.

Solution (ii)

\[ \begin{aligned} (x + 2)^3 &= x^3 - 4 \\ x^3 + 6x^2 + 12x + 8 &= x^3 - 4 \\ x^3 + 6x^2 + 12x + 8 - x^3 + 4 &= 0 \\ 6x^2 + 12x + 12 &= 0 \end{aligned} \]

The highest power of \(x\) is 2. Therefore, the equation reduces to quadratic form:
\(6x^2 + 12x + 12 = 0\)

Hence, it is also a quadratic equation.

Transformation Insight

Even if the original equation contains higher powers, cancellation may reduce it to a quadratic equation.

Key Learning

• Always simplify the equation fully before deciding its type
• Degree is determined after simplification, not from initial appearance
• Expansion and cancellation can reduce higher-degree equations to quadratic

Common Mistakes

• Incorrect expansion of \((a+b)^3\)
• Forgetting to move all terms to one side
• Judging degree before simplification

Find the roots of the equation \(2x^2 - 5x + 3 = 0\) using factorisation.

Concept Used

In factorisation, we split the middle term \(bx\) into two terms such that:

• Product = \(a \times c\)
• Sum = coefficient of \(x\)

Solution

Given:
\(2x^2 - 5x + 3 = 0\)

Compute:
\(a \times c = 2 \times 3 = 6\)

We need two numbers whose:

• Product = \(+6\)
• Sum = \(-5\)

These numbers are \(-2\) and \(-3\)

\[ \begin{aligned} 2x^2 - 5x + 3 &= 0 \\ 2x^2 - 2x - 3x + 3 &= 0 \\ 2x(x - 1) - 3(x - 1) &= 0 \\ (x - 1)(2x - 3) &= 0 \end{aligned} \]

Using zero-product property:

\[ \begin{aligned} x - 1 &= 0 \Rightarrow x = 1 \\ 2x - 3 &= 0 \Rightarrow x = \frac{3}{2} \end{aligned} \]

Roots: \(x = 1,\ \frac{3}{2}\)

Verification (Recommended in Exams)

Substituting \(x = 1\):
\(2(1)^2 - 5(1) + 3 = 2 - 5 + 3 = 0\)

Substituting \(x = \frac{3}{2}\):
\(2\left(\frac{3}{2}\right)^2 - 5\left(\frac{3}{2}\right) + 3 = 0\)

Both values satisfy the equation.

Factorisation Flow

Key Learning

• Always consider signs while splitting middle term
• If \(b\) is negative → both split terms are negative
• Factorisation is fastest when roots are rational

Common Mistakes

• Ignoring sign of middle term
• Wrong pair selection for \(ac\)
• Incorrect grouping

Find the roots of the quadratic equation \(6x^2 - x - 2 = 0\).

Concept Used

Split the middle term such that:

• Product = \(a \times c = 6 \times (-2) = -12\)
• Sum = coefficient of \(x = -1\)

Solution

Given:
\(6x^2 - x - 2 = 0\)

We need two numbers whose:
Product = \(-12\), Sum = \(-1\)

These numbers are \(-4\) and \(+3\)

\[ \begin{aligned} 6x^2 - x - 2 &= 0 \\ 6x^2 - 4x + 3x - 2 &= 0 \\ 2x(3x - 2) + 1(3x - 2) &= 0 \\ (3x - 2)(2x + 1) &= 0 \end{aligned} \]

Using zero-product property:

\[ \begin{aligned} 3x - 2 &= 0 \Rightarrow x = \frac{2}{3} \\ 2x + 1 &= 0 \Rightarrow x = -\frac{1}{2} \end{aligned} \]

Roots: \(x = \frac{2}{3},\ -\frac{1}{2}\)

Verification

Substituting \(x = \frac{2}{3}\):
\(6\left(\frac{2}{3}\right)^2 - \frac{2}{3} - 2 = 0\)

Substituting \(x = -\frac{1}{2}\):
\(6\left(-\frac{1}{2}\right)^2 + \frac{1}{2} - 2 = 0\)

Factorisation Flow

Key Learning

• If \(ac\) is negative → split terms will have opposite signs
• Always verify sum matches \(b\)
• Reordering terms can simplify grouping

Common Mistakes

• Wrong sign selection while splitting
• Incorrect grouping of terms
• Final answer sign error (very common)

Find the roots of the quadratic equation \(3x^2 - 2\sqrt{6}x + 2 = 0\).

Concept Used

Identify the identity:
\(a^2 - 2ab + b^2 = (a - b)^2\)

Solution

\[ \begin{aligned} 3x^2 - 2\sqrt{6}x + 2 &= 0 \\ (\sqrt{3}x)^2 - 2(\sqrt{3}x)(\sqrt{2}) + (\sqrt{2})^2 &= 0 \\ (\sqrt{3}x - \sqrt{2})^2 &= 0 \end{aligned} \]

Taking square root:
\[ \sqrt{3}x - \sqrt{2} = 0 \]

\[ \begin{aligned} \sqrt{3}x &= \sqrt{2} \\ x &= \frac{\sqrt{2}}{\sqrt{3}} \end{aligned} \]

Rationalising:
\[ x = \sqrt{\frac{2}{3}} \]

Roots: \(x = \sqrt{\frac{2}{3}},\ \sqrt{\frac{2}{3}}\)
(Equal roots)

Perfect Square Pattern Recognition

Key Learning

• If expression matches \(a^2 - 2ab + b^2\), use identity directly
• Such equations always give equal roots
• Recognising patterns saves time over full methods

Board Insight

• Frequently asked in 2–3 mark questions
• Tests identity recognition + simplification skills
• Often combined with discriminant concept (D = 0)

Common Mistakes

• Not identifying perfect square pattern
• Incorrect handling of square roots
• Not rationalising final answer (presentation marks)

Find the discriminant of the quadratic equation \(2x^2 - 4x + 3 = 0\), and hence determine the nature of its roots.

Concept Used

Discriminant:
\(D = b^2 - 4ac\)

Solution

Given:
\(2x^2 - 4x + 3 = 0\)

Comparing with \(ax^2 + bx + c = 0\):
\(a = 2,\ b = -4,\ c = 3\)

\[ \begin{aligned} D &= b^2 - 4ac \\ &= (-4)^2 - 4(2)(3) \\ &= 16 - 24 \\ &= -8 \end{aligned} \]

Since \(D < 0\), the equation has:

• No real roots
• Two complex (non-real) conjugate roots

Graphical Insight

The parabola does not intersect the x-axis, confirming there are no real roots.

Advanced Insight

When \(D < 0\), roots are of the form:
\(x = \frac{-b \pm i\sqrt{|D|}}{2a}\)

Key Learning

• Discriminant directly determines nature of roots
• Negative \(D\) implies complex roots (not visible on graph)
• No need to solve fully if only nature is asked

Common Mistakes

• Sign error in \(b^2\)
• Incorrect multiplication in \(4ac\)
• Writing "no solution" instead of "no real roots"

Find the discriminant of the equation \(3x^2 - 2x + \frac{1}{3} = 0\), determine the nature of its roots, and find them if they are real.

Concept Used

• Discriminant: \(D = b^2 - 4ac\)
• If \(D = 0\) → roots are real and equal

Solution (Method 1: Eliminating Fraction)

Given:
\(3x^2 - 2x + \frac{1}{3} = 0\)

Multiply both sides by 3:
\(9x^2 - 6x + 1 = 0\)

Here, \(a = 9,\ b = -6,\ c = 1\)

\[ \begin{aligned} D &= b^2 - 4ac \\ &= (-6)^2 - 4(9)(1) \\ &= 36 - 36 \\ &= 0 \end{aligned} \]

Since \(D = 0\), the roots are real and equal.

\[ x = \frac{-b}{2a} = \frac{-(-6)}{2 \times 9} = \frac{6}{18} = \frac{1}{3} \]

Roots: \(x = \frac{1}{3},\ \frac{1}{3}\)

Alternative Method (Direct without Scaling)

Compare directly:
\(a = 3,\ b = -2,\ c = \frac{1}{3}\)

\[ \begin{aligned} D &= (-2)^2 - 4(3)\left(\frac{1}{3}\right) \\ &= 4 - 4 \\ &= 0 \end{aligned} \]

Again, \(D = 0\) ⇒ equal roots.

Graphical Insight

The parabola touches the x-axis at exactly one point, confirming equal roots.

Key Learning

• Multiplying to remove fractions simplifies calculations
• If \(D = 0\), equation becomes a perfect square
• Equal roots always correspond to a tangent to x-axis

Common Mistakes

• Ignoring fractions and making calculation errors
• Wrong value of \(a, b, c\)
• Not simplifying final answer

Relationship Between Roots and Coefficients

If \( \alpha \) and \( \beta \) are roots of \(ax^2 + bx + c = 0\), then:

\( \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \)

These formulas allow solving problems without actually finding roots.

Applications

• Forming quadratic equation from given roots
• Finding new equations using transformations
• Solving symmetric expressions

Forming a Quadratic Equation from Given Roots

If roots are \( \alpha \) and \( \beta \), then equation is:

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

If coefficient of \(x^2\) is not 1, multiply entire equation accordingly.

Conditions for Nature of Roots (Without Solving)

• \(b^2 - 4ac > 0\) → real and distinct
• \(b^2 - 4ac = 0\) → real and equal
• \(b^2 - 4ac < 0\) → complex roots

These are frequently used in case-study and assertion-reason questions.

Parabola Insights (Deep Understanding)

• If \(a > 0\) → parabola opens upward (minimum value exists)
• If \(a < 0\) → parabola opens downward (maximum value exists)
• Vertex gives optimal value (important in application problems)

Vertex formula:
\(x = -\frac{b}{2a}\)

Maximum and Minimum Value

For \(ax^2 + bx + c\), the extreme value occurs at:

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

• If \(a > 0\) → minimum value
• If \(a < 0\) → maximum value

This concept is widely used in optimization problems.

Condition for Equal Roots

For equal roots:
\(b^2 - 4ac = 0\)

Such equations can always be written as:
\((x - \alpha)^2 = 0\)

Transformation of Roots

If roots are modified (e.g., \( \alpha + k, \beta + k \)), new equation can be formed without solving original roots.

This concept is highly important for higher classes and Olympiad-level questions.

Hidden Quadratic Forms

Some equations are not directly quadratic but can be converted:

• \(x + \frac{1}{x} = k\)
• Substitute \(x = y + \frac{1}{y}\)
• Leads to quadratic in new variable

These are common in competitive exams.

Roots and Graph Relationship

• Roots = x-intercepts
• No intersection → complex roots
• Tangent → equal roots

High-Speed Exam Shortcuts

• If sum & product given → use root formulas
• If D asked → no need to solve fully
• If expression symmetric → use identities
• If coefficients large → directly use formula

Ultimate Mistake Checklist

• Sign error in \(b^2 - 4ac\)
• Incorrect factorisation
• Ignoring extraneous roots in word problems
• Not simplifying final answer
• Wrong interpretation of discriminant

Master Insight (Top 1% Understanding)

Every quadratic equation represents a parabola. Solving the equation means finding where this parabola intersects the x-axis.

Understanding this connection between algebra and geometry is the key to mastering the chapter.