Introduction

Chapter 4: Quadratic Equations

Page 1 of 5: Introduction to Quadratic Equations

{{FORMULA: expr=ax² + bx + c = 0 | symbols=a:leading coefficient, b:linear coefficient, c:constant}}

Concept Introduction

Welcome to the world of quadratic equations! In your previous chapter on polynomials, you became familiar with the quadratic polynomial, an expression of the form ax² + bx + c, where a ≠ 0. What happens when we take this expression and set it equal to a value, specifically zero? We create an equation, and that's the birth of a quadratic equation.

These equations are more than just abstract algebra; they are powerful tools for modeling the world around us. Imagine a charity trust planning to build a rectangular prayer hall with a specific carpet area of 300 square metres. They have a design constraint: the length must be one metre more than twice its breadth.

If we let the breadth be x metres, the length becomes (2x + 1) metres. The area of a rectangle is length × breadth, so we get:

Area = (2x + 1) × x = 2x² + x

Since the required area is 300 m², we can set up the equation:

2x² + x = 300

By rearranging it, we get 2x² + x - 300 = 0. This is a classic quadratic equation. Solving it would give us the exact breadth x needed for the hall. This chapter will teach you how to set up and solve such equations.

A quick historical note: While often credited to later mathematicians, the Babylonians were solving problems equivalent to quadratic equations thousands of years ago! Ancient Indian mathematicians like Brahmagupta and Sridharacharya also made monumental contributions, developing formulas to solve them.

Definitions & Formulas

Understanding the language of quadratics is the first step. Here are the core terms you'll encounter.

Derivation: From Word Problem to Standard Form

The first critical skill is translating a real-world problem into a mathematical equation. Let's break down the logic using an example from your textbook.

Problem: John and Jivanti together have 45 marbles. Both lose 5 marbles each, and the product of the marbles they now have is 124. How do we represent this situation mathematically?

1. Assign a variable. Let the number of marbles John had initially be x.

2. Express other quantities in terms of the variable. Since they have 45 marbles in total, the number of marbles Jivanti had is 45 - x.

3. Apply the conditions of the problem. Both lose 5 marbles.

   • Marbles left with John = x - 5
   • Marbles left with Jivanti = (45 - x) - 5 = 40 - x

4. Formulate the equation based on the given relationship. The product of the marbles they now have is 124.

   (x - 5)(40 - x) = 124
   

5. Simplify the equation by expanding the product. We use the distributive property (or FOIL method).

   x(40) + x(-x) - 5(40) - 5(-x) = 124
   

   40x - x² - 200 + 5x = 124
   

   -x² + 45x - 200 = 124
   

6. Rearrange the equation into the standard form ax² + bx + c = 0. To do this, we move all terms to one side, making the other side zero. It's conventional to make the x² term positive.

   -x² + 45x - 200 - 124 = 0
   

   -x² + 45x - 324 = 0
   

   Now, multiply the entire equation by -1 to make the x² term positive.

   x² - 45x + 324 = 0
   

   This is the final quadratic equation representing the situation.

{{KEY: type=concept | title=The Standard Form is Your North Star | text=Always simplify and rearrange an equation into the standard form ax² + bx + c = 0 before trying to classify or solve it. The single most important condition is that 'a' (the coefficient of x²) must not be zero. If a = 0, the equation becomes a linear equation (bx + c = 0), not a quadratic one.}}

Solved Examples

Let's practice identifying quadratic equations. The key is to simplify the given equation and check if it can be written in the form ax² + bx + c = 0 where a ≠ 0.

Example 1: Simple Rearrangement (Easy)

Given: x(2x + 3) = x² + 1

To Find: Check if the equation is a quadratic equation.

Solution:

1. First, expand the left-hand side (LHS) of the equation.

   LHS = 2x² + 3x
   

2. Now, set the expanded LHS equal to the right-hand side (RHS).

   2x² + 3x = x² + 1
   

3. Move all terms from the RHS to the LHS to get the standard form. Remember to change their signs.

   2x² - x² + 3x - 1 = 0
   

4. Combine like terms to get the final simplified form.

   x² + 3x - 1 = 0
   

5. Compare this to the standard form ax² + bx + c = 0. Here, a = 1, b = 3, c = -1. Since a ≠ 0, it is a quadratic equation.

Final Answer: Yes, the given equation is a quadratic equation.

Example 2: Expansion of Squares (Medium)

Given: (x - 2)² + 1 = 2x - 3

To Find: Check if the equation is a quadratic equation.

Solution:

1. Expand the squared term (x - 2)² on the LHS using the identity (a - b)² = a² - 2ab + b².

   LHS = (x² - 4x + 4) + 1 = x² - 4x + 5
   

2. Set the simplified LHS equal to the RHS.

   x² - 4x + 5 = 2x - 3
   

3. Rearrange all terms to one side to get the standard form.

   x² - 4x - 2x + 5 + 3 = 0
   

4. Combine like terms.

   x² - 6x + 8 = 0
   

5. This equation is in the form ax² + bx + c = 0, with a = 1, b = -6, c = 8. Since a ≠ 0, the condition is met.

Final Answer: Yes, the given equation is a quadratic equation.

Example 3: The Disappearing x² Term (Hard)

Given: x(x + 1) + 8 = (x + 2)(x - 2)

To Find: Check if the equation is a quadratic equation.

Solution:

1. Simplify the LHS by expanding x(x + 1).

   LHS = x² + x + 8
   

2. Simplify the RHS using the identity (a + b)(a - b) = a² - b².

   RHS = x² - 2² = x² - 4
   

3. Now, equate the simplified LHS and RHS.

   x² + x + 8 = x² - 4
   

4. Move all terms to one side.

   x² - x² + x + 8 + 4 = 0
   

5. Combine like terms. Notice what happens to the x² term.

   0x² + x + 12 = 0
   

   Which simplifies to:

   x + 12 = 0
   

6. This final equation is of the form bx + c = 0. The term with x² has vanished, meaning a = 0. Therefore, it is not a quadratic equation; it is a linear equation.

Final Answer: No, the given equation is not a quadratic equation.

Example 4: The Cancelling x³ Term (Tricky)

Given: (x + 2)³ = x³ - 4

To Find: Check if the equation is a quadratic equation.

Solution:

1. This equation appears to be cubic because of the (x+2)³ and x³ terms. Let's expand the LHS using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³.

   LHS = x³ + 3(x²)(2) + 3(x)(2²) + 2³
   

   LHS = x³ + 6x² + 12x + 8
   

2. Now, set the expanded LHS equal to the RHS.

   x³ + 6x² + 12x + 8 = x³ - 4
   

3. Rearrange the terms to bring everything to one side.

   x³ - x³ + 6x² + 12x + 8 + 4 = 0
   

4. Combine like terms. The x³ terms cancel each other out.

   6x² + 12x + 12 = 0
   

5. We can simplify this equation by dividing all terms by 6.

   x² + 2x + 2 = 0
   

6. This is in the standard form ax² + bx + c = 0 where a = 1, b = 2, c = 2. Since a ≠ 0, the equation is indeed quadratic.

Final Answer: Yes, the given equation is a quadratic equation.

Tips & Tricks

Use these shortcuts to quickly identify and form quadratic equations.

Common Mistakes

Here are some common pitfalls to avoid when working with quadratic equations.

Brain-Teaser Questions

Test your understanding with these challenging problems.

1. The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. Formulate the quadratic equation to find the length and breadth of the plot.

💡 Answer: Let the breadth be x metres. Then the length is (2x + 1) metres. Area = x(2x + 1) = 528. This simplifies to 2x² + x = 528, and the standard form is 2x² + x - 528 = 0.

2. For what value of k does the equation (k - 12)x² + 2(k - 12)x + 2 = 0 fail to be a quadratic equation?

💡 Answer: An equation fails to be quadratic if the coefficient of the x² term is zero. Here, the coefficient is (k - 12). Setting it to zero: k - 12 = 0 → k = 12. If k=12, the equation becomes 0x² + 0x + 2 = 0, which is 2 = 0, a false statement. Thus, for k=12, it is not a quadratic equation.

3. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Formulate the quadratic equation to find the speed of the train.

💡 Answer: Let the uniform speed be x km/h. Time taken = Distance / Speed = 480 / x. New speed = (x - 8) km/h. New time taken = 480 / (x - 8). The new time is 3 hours more than the original time: (480 / (x - 8)) - (480 / x) = 3. Multiplying by x(x - 8) to clear denominators gives 480x - 480(x - 8) = 3x(x - 8). 480x - 480x + 3840 = 3x² - 24x. 3x² - 24x - 3840 = 0. Dividing by 3 gives the final equation: x² - 8x - 1280 = 0.

Mini Cheatsheet

Screenshot this table for a quick last-minute revision of today's concepts.

Quadratic Equations

Chapter 4: Quadratic Equations (Page 2/5)

4.2 What is a Quadratic Equation?

In the previous chapter on Polynomials, you studied quadratic polynomials. A quadratic polynomial is a polynomial of degree 2, with the general form ax² + bx + c, where a, b, and c are real numbers and, most importantly, a ≠ 0.

Now, what happens when we set this polynomial equal to zero? When we equate a quadratic polynomial to zero, we get a quadratic equation.

Definition: A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.

This form, ax² + bx + c = 0, is called the standard form of a quadratic equation.

• x is the variable.
• a is the coefficient of the x² term.
• b is the coefficient of the x term.
• c is the constant term.

Why is the condition a ≠ 0 so important? If a were equal to 0, the term ax² would become 0 * x² = 0, and the equation would simplify to bx + c = 0. This is a linear equation, not a quadratic one. The x² term is what defines an equation as quadratic.

Let's look at the example from the introduction: A prayer hall has a carpet area of 300 m². Its length is one metre more than twice its breadth. Let the breadth be x metres. Then, the length is (2x + 1) metres. Area = Length × Breadth 300 = (2x + 1) * x 300 = 2x² + x Rearranging this to the standard form (by moving all terms to one side), we get: 2x² + x – 300 = 0 This is a quadratic equation where a = 2, b = 1, and c = -300. Solving this equation will give us the value of the breadth, x.

Checking if an Equation is Quadratic

Sometimes, an equation might not look like a quadratic equation at first glance. It is essential to simplify the equation completely and rearrange it into the standard form ax² + bx + c = 0 before making a conclusion.

Steps to Check if an Equation is Quadratic:

1. Eliminate any brackets by expanding the terms. Use algebraic identities like (a + b)² = a² + 2ab + b² or (a + b)³ = a³ + 3a²b + 3ab² + b³ where needed.
2. Bring all terms to the left-hand side (LHS) of the equation, leaving 0 on the right-hand side (RHS).
3. Combine like terms (all x² terms together, all x terms together, and all constant terms together).
4. Check the highest power (degree) of the variable x in the simplified equation.
5. If the degree is 2, the equation is a quadratic equation. If the degree is not 2, it is not a quadratic equation.

Worked Example 1: Check whether the following are quadratic equations.

(i) (x – 2)² + 1 = 2x – 3

• Step 1: Expand the LHS. Using the identity (a – b)² = a² – 2ab + b², we get: (x² – 2x2 + 2²) + 1 = 2x – 3 x² – 4x + 4 + 1 = 2x – 3 x² – 4x + 5 = 2x – 3

• Step 2 & 3: Bring all terms to LHS and simplify. x² – 4x + 5 – 2x + 3 = 0 x² + (–4x – 2x) + (5 + 3) = 0 x² – 6x + 8 = 0

• Step 4 & 5: Check the degree. The highest power of x is 2. The equation is in the form ax² + bx + c = 0 (with a = 1, b = -6, c = 8). Answer: Yes, the given equation is a quadratic equation.

(ii) x(x + 1) + 8 = (x + 2)(x – 2)

• Step 1: Expand both sides. LHS: x(x + 1) + 8 = x² + x + 8 RHS: (x + 2)(x – 2) = x² – 2² = x² – 4 (using the identity (a + b)(a – b) = a² – b²)

• Step 2 & 3: Rewrite the equation and simplify. x² + x + 8 = x² – 4 x² + x + 8 – x² + 4 = 0 (x² – x²) + x + (8 + 4) = 0 0 + x + 12 = 0 x + 12 = 0

• Step 4 & 5: Check the degree. The highest power of x is 1. This is a linear equation. Answer: No, the given equation is not a quadratic equation.

(iii) (x + 2)³ = x³ – 4

• Step 1: Expand the LHS. Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³, we get: x³ + 3(x²)(2) + 3(x)(2²) + 2³ = x³ – 4 x³ + 6x² + 12x + 8 = x³ – 4

• Step 2 & 3: Bring all terms to LHS and simplify. x³ + 6x² + 12x + 8 – x³ + 4 = 0 (x³ – x³) + 6x² + 12x + (8 + 4) = 0 6x² + 12x + 12 = 0

• Step 4 & 5: Check the degree. The highest power of x is 2. We can simplify this further by dividing by 6: x² + 2x + 2 = 0. This is in the form ax² + bx + c = 0. Answer: Yes, the given equation is a quadratic equation.

Teacher's Note: Be careful! Example (ii) appeared to be quadratic because of the x(x+1) term, but it simplified to a linear equation. Example (iii) appeared to be cubic because of the x³ term, but it simplified to a quadratic equation. Always simplify completely before deciding.

Representing Real-Life Situations as Quadratic Equations

A key skill in mathematics is translating word problems into the language of algebra. Many real-world problems involving area, distance, speed, time, and numbers can be modeled using quadratic equations.

Steps to Formulate a Quadratic Equation from a Word Problem:

1. Read the problem carefully to understand the given information and what needs to be found.
2. Identify the unknown quantity and represent it with a variable, usually x.
3. Use the information in the problem to express all other unknown quantities in terms of x.
4. Form an equation that relates these quantities based on the conditions given in the problem.
5. Simplify the equation and write it in the standard form ax² + bx + c = 0.

Worked Example 2: Represent the following situations in the form of quadratic equations.

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We want to find out how many marbles they had to start with.

• Step 1 & 2: Identify and represent the unknown. Let the number of marbles John had be x.

• Step 3: Express other quantities in terms of x. Since they have 45 marbles together, the number of marbles Jivanti had = 45 – x. After losing 5 marbles, John has x – 5 marbles. After losing 5 marbles, Jivanti has (45 – x) – 5 = 40 – x marbles.

• Step 4: Form the equation based on the given condition. The problem states that the product of the marbles they now have is 124. So, (Number of John's marbles) × (Number of Jivanti's marbles) = 124 (x – 5)(40 – x) = 124

• Step 5: Simplify to standard form. x(40 – x) – 5(40 – x) = 124 40x – x² – 200 + 5x = 124 –x² + 45x – 200 = 124 –x² + 45x – 200 – 124 = 0 –x² + 45x – 324 = 0 To make the x² term positive (a standard convention), multiply the entire equation by -1: x² – 45x + 324 = 0

Answer: The situation is represented by the quadratic equation x² – 45x + 324 = 0, where x is the number of marbles John initially had.

Solution of a Quadratic Equation by Factorisation — Part 1

{{FORMULA: expr=ax² + bx + c = 0 | symbols=a:leading coefficient, b:linear coefficient, c:constant}}

Solution of a Quadratic Equation by Factorisation — Part 1

Welcome back! In our previous lesson, we learned how to represent real-life situations as quadratic equations. But what's the use of an equation if you can't solve it? Finding the 'solution' to a problem is the ultimate goal.

Imagine an architect designing a rectangular community hall. The client has a specific requirement: the length must be one meter more than twice its breadth, and the total floor area must be exactly 300 square meters. How does the architect find the exact dimensions? This is where solving quadratic equations becomes a crucial real-world skill. The architect would set up the equation x(2x + 1) = 300, which simplifies to 2x² + x – 300 = 0. The value of x that makes this equation true is the answer they need.

In this section, we'll dive into our first powerful method for solving these equations: Factorisation. It's a technique that breaks down a complex quadratic expression into simpler parts, making the solution easy to find.

Definitions & Formulas

Before we start solving, let's be crystal clear on the terminology. These terms are the building blocks of our entire chapter.

{{KEY: type=concept | title=Roots vs. Zeroes | text=The roots of the quadratic equation ax² + bx + c = 0 are the exact same values as the zeroes of the quadratic polynomial ax² + bx + c. The terms are often used interchangeably.}}

The Logic of Factorisation: The Zero-Product Rule

How does breaking an expression into factors magically give us the solution? The secret lies in a simple but powerful idea called the Zero-Product Rule. Let's break down the logic.

1. The Goal: Our starting point is a standard quadratic equation.

   ax² + bx + c = 0
   

2. The Transformation: The method of factorisation aims to transform the quadratic polynomial on the left-hand side (LHS) into a product of two linear factors.

   (Some linear expression) × (Another linear expression) = 0
   

   For example, we might turn 2x² – 5x + 3 into (2x – 3)(x – 1).

3. Applying the Rule: Now our equation looks like this:

   (2x – 3)(x – 1) = 0
   

   Think about it: if you multiply two numbers and the result is 0, what can you say about those numbers? At least one of them must be 0. This is the Zero-Product Rule.

4. Finding the Roots: Applying this rule, we can set each factor equal to zero independently.

   Either  2x – 3 = 0   OR   x – 1 = 0
   

5. The Final Step: We now have two very simple linear equations. Solving them gives us the roots of the original, more complex quadratic equation.

   If 2x – 3 = 0, then x = 3/2
   

   If x – 1 = 0, then x = 1
   

   And there we have it! The roots are 3/2 and 1. This logical flow is the backbone of the factorisation method.

Solved Examples

Let's put this theory into practice. We'll work through examples from the NCERT textbook, starting with the basics and moving to more challenging problems.

Example 1: Basic Factorisation

Given: The quadratic equation 2x² – 5x + 3 = 0.

To Find: The roots of the equation by factorisation.

Solution:

1. Our goal is to split the middle term, –5x. We need two numbers that add up to –5 and multiply to a × c = 2 × 3 = 6. The numbers –2 and –3 work perfectly (–2 + –3 = –5 and –2 × –3 = 6).

2. Rewrite the equation by splitting the middle term.

   2x² – 2x – 3x + 3 = 0
   

3. Factor by grouping. Take out the common factor from the first two terms and the last two terms.

   2x(x – 1) – 3(x – 1) = 0
   

   Notice that (x – 1) is now a common factor.

4. Factor out the common bracket (x – 1).

   (2x – 3)(x – 1) = 0
   

5. Apply the Zero-Product Rule. Set each factor to zero and solve for x.

   2x – 3 = 0   gives   x = 3/2
   

   x – 1 = 0   gives   x = 1
   

Final Answer:

The roots of the equation are x = 3/2 and x = 1.

Example 2: Factorisation with Negative Terms

Given: The quadratic equation 6x² – x – 2 = 0.

To Find: The roots of the equation.

Solution:

1. Identify the coefficients: a = 6, b = –1, c = –2. We need two numbers that sum to b = –1 and have a product of a × c = 6 × (–2) = –12.

2. Let's think of factors of –12: (1, -12), (2, -6), (3, -4), (4, -3)... The pair 3 and –4 sums to –1.

3. Split the middle term –x into +3x – 4x.

   6x² + 3x – 4x – 2 = 0
   

4. Factor by grouping.

   3x(2x + 1) – 2(2x + 1) = 0
   

   Here, (2x + 1) is the common factor.

5. Factor out the common bracket.

   (3x – 2)(2x + 1) = 0
   

6. Apply the Zero-Product Rule and solve for x.

   3x – 2 = 0   gives   x = 2/3
   

   2x + 1 = 0   gives   x = –1/2
   

Final Answer:

The roots of the equation are x = 2/3 and x = –1/2.

Example 3: Factorisation with Irrational Numbers (Surds)

Given: The quadratic equation 3x² – 2√6x + 2 = 0.

To Find: The roots of the equation.

Solution:

1. This looks tricky, but the process is the same. We need two numbers that sum to –2√6 and multiply to a × c = 3 × 2 = 6.

2. How can we get √6? The number we're looking for must involve √6. Let's try splitting –2√6x into –√6x – √6x. Sum: (–√6) + (–√6) = –2√6. Product: (–√6) × (–√6) = 6. It works!

3. Rewrite the equation with the split middle term.

   3x² – √6x – √6x + 2 = 0
   

4. Factoring this requires a small trick. Remember that 3 = √3 × √3 and √6 = √3 × √2.

   (√3x × √3x) – (√3 × √2)x – (√3 × √2)x + (√2 × √2) = 0
   

5. Now, factor by grouping. From the first two terms, we can take out √3x. From the last two, we can take out –√2.

   √3x(√3x – √2) – √2(√3x – √2) = 0
   

6. Factor out the common bracket (√3x – √2).

   (√3x – √2)(√3x – √2) = 0
   

7. In this case, both factors are identical. We only need to solve one.

   √3x – √2 = 0   gives   √3x = √2   gives   x = √2/√3
   

8. To rationalize the denominator, we can also write x as √6/3. Since the factor is repeated, the root is also repeated.

Final Answer:

The equation has two equal roots: x = √2/√3 and x = √2/√3.

Example 4: Application in a Word Problem (Tricky)

Given: A rectangular prayer hall's breadth is x meters. Its length is 2x + 1 meters. The total area is 300 m². This situation is modeled by the equation 2x² + x – 300 = 0.

To Find: The dimensions (breadth and length) of the prayer hall.

Solution:

1. We need to find the roots of the equation 2x² + x – 300 = 0. First, we find the product a × c = 2 × (–300) = –600.

2. We need two numbers that multiply to –600 and have a difference of b = +1. This is a bit of a search. Let's list factors of 600 near its square root (which is about 24.5). 20 × 30 ... difference is 10. 24 × 25 ... difference is 1. This is our pair! Since the sum is +1, the numbers are +25 and –24.

3. Split the middle term +x into +25x – 24x.

   2x² + 25x – 24x – 300 = 0
   

4. Factor by grouping. This arrangement is tricky. Let's rearrange for easier factoring: 2x² – 24x + 25x – 300 = 0.

   2x(x – 12) + 25(x – 12) = 0
   

5. Factor out the common bracket (x – 12).

   (2x + 25)(x – 12) = 0
   

6. Apply the Zero-Product Rule to find the possible values of x.

   2x + 25 = 0   gives   x = –25/2 = –12.5
   

   x – 12 = 0   gives   x = 12
   

7. Interpret the result. The variable x represents the breadth of the hall. A physical dimension like breadth cannot be negative. Therefore, we must discard the root x = –12.5.

8. The only valid solution is x = 12. Now we can find the dimensions.

   • Breadth = x = 12 meters.
   • Length = 2x + 1 = 2(12) + 1 = 25 meters.

Final Answer:

The breadth of the prayer hall is 12 m and the length is 25 m.

Tips & Tricks

Mastering factorisation is about practice and recognizing patterns. Here are some tricks to speed up your process.

Common Mistakes

Factorisation is precise work, and small errors can lead to wrong answers. Here are the most common pitfalls to watch out for.

Brain-Teaser Questions

Ready to test your skills? These questions require careful application of the concepts we've learned.

1. Find the roots of the equation 2x² – x + ⅛ = 0 by factorisation.

   💡 Answer: First, multiply the entire equation by 8 to eliminate the fraction: 16x² – 8x + 1 = 0. We need two numbers that sum to –8 and multiply to 16×1 = 16. The numbers are –4 and –4. 16x² – 4x – 4x + 1 = 0 4x(4x – 1) – 1(4x – 1) = 0 (4x – 1)(4x – 1) = 0 The equation has two equal roots: x = 1/4.

2. Find two consecutive positive integers, the sum of whose squares is 365.

   💡 Answer: Let the integers be x and x + 1. The equation is x² + (x + 1)² = 365. x² + x² + 2x + 1 = 365 2x² + 2x – 364 = 0 Divide by 2: x² + x – 182 = 0. We need two numbers that multiply to –182 and have a difference of 1. The numbers are 14 and –13. (x + 14)(x – 13) = 0. The roots are x = –14 and x = 13. Since we need positive integers, we discard –14. The integers are x = 13 and x + 1 = 14.

3. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the lengths of the base and altitude.

   💡 Answer: Let the base be x cm. Then the altitude is (x – 7) cm. By Pythagoras' theorem: base² + altitude² = hypotenuse². x² + (x – 7)² = 13² x² + x² – 14x + 49 = 169 2x² – 14x – 120 = 0 Divide by 2: x² – 7x – 60 = 0. We need two numbers that multiply to –60 and sum to –7. The numbers are –12 and 5. (x – 12)(x + 5) = 0. The roots are x = 12 and x = –5. Since length cannot be negative, we discard –5. The base is x = 12 cm. The altitude is x – 7 = 12 – 7 = 5 cm.

Mini Cheatsheet

Here's a quick summary of today's key points. Screenshot this for your last-minute revision!

Solution of a Quadratic Equation by Factorisation — Part 2

Chapter 4: Quadratic Equations

Page 4 of 5: Nature of Roots

4.4 Nature of Roots

In our previous lesson, we learned the powerful quadratic formula to find the roots of any quadratic equation of the form ax² + bx + c = 0.

The formula is: x = [-b ± √(b² - 4ac)] / 2a

Have you ever wondered why some quadratic equations have two different solutions, some have only one, and some have no solution at all? The answer lies inside the square root symbol.

The expression b² - 4ac is the key. It "discriminates" or determines the type of roots the equation will have. For this reason, it is called the discriminant of the quadratic equation.

Definition: Discriminant For a quadratic equation ax² + bx + c = 0, the discriminant, denoted by D, is given by the expression: D = b² - 4ac

The value of the discriminant tells us about the nature of the roots without actually solving the equation. Let's explore the three possible cases.

Case 1: Discriminant is positive (D > 0)

If b² - 4ac > 0, the value inside the square root is a positive number. The quadratic formula becomes x = [-b ± √(a positive number)] / 2a. The square root of a positive number is a real, non-zero value. Because of the ± sign, we will get two different values for x:

1. x₁ = [-b + √(b² - 4ac)] / 2a
2. x₂ = [-b - √(b² - 4ac)] / 2a

Conclusion: When D > 0, the quadratic equation has two distinct real roots.

Case 2: Discriminant is zero (D = 0)

If b² - 4ac = 0, the value inside the square root is zero. The quadratic formula becomes x = [-b ± √0] / 2a, which simplifies to x = [-b ± 0] / 2a. Adding or subtracting zero does not change the value. So, both roots are the same:

1. x₁ = -b / 2a
2. x₂ = -b / 2a

Conclusion: When D = 0, the quadratic equation has two equal real roots (also called coincident roots).

Case 3: Discriminant is negative (D < 0)

If b² - 4ac < 0, the value inside the square root is a negative number. The quadratic formula becomes x = [-b ± √(a negative number)] / 2a. In the system of real numbers, we cannot find the square root of a negative number. For example, there is no real number that, when squared, gives -4.

Conclusion: When D < 0, the quadratic equation has no real roots.

Summary Table: Nature of Roots

Worked Examples

Let's apply this knowledge to some problems.

Example 1 (Illustrating D < 0)

Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0 and hence find the nature of its roots.

Solution:

1. Identify coefficients: The given equation is 2x² – 4x + 3 = 0. Comparing this with the standard form ax² + bx + c = 0, we get:

   • a = 2
   • b = -4
   • c = 3

2. Calculate the discriminant (D): D = b² - 4ac D = (-4)² - 4(2)(3) D = 16 - 24 D = -8

3. Determine the nature of roots: Since D = -8, which is less than 0 (D < 0), the equation falls into Case 3.

   Answer: The discriminant is -8. Therefore, the given quadratic equation has no real roots.

Example 2 (Illustrating D = 0)

Find the discriminant of the equation 3x² – 2x + 1/3 = 0. Find the nature of its roots. If they are real, find them.

Solution:

1. Identify coefficients: The given equation is 3x² – 2x + 1/3 = 0.

   • a = 3
   • b = -2
   • c = 1/3

2. Calculate the discriminant (D): D = b² - 4ac D = (-2)² - 4(3)(1/3) D = 4 - 4 D = 0

3. Determine the nature of roots: Since D = 0, the equation falls into Case 2. The equation has two equal real roots.

4. Find the roots: Since the roots are real and equal, we can find them using the simplified formula x = -b / 2a. x = -(-2) / (2 * 3) x = 2 / 6 x = 1/3

   Answer: The discriminant is 0. The equation has two equal real roots, which are 1/3 and 1/3.

Example 3 (Application Problem with D > 0)

A pole has to be erected on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is this situation possible? If yes, at what distances from the two gates should the pole be erected?

Solution:

1. Formulate the problem mathematically: Let the position of the pole be P. Let the distance from gate B to the pole, BP, be x metres. The problem states that the difference in distances from gates A and B is 7 m. So, AP - BP = 7, which means AP = BP + 7 = x + 7 metres. The gates A and B are diametrically opposite, so the line AB is the diameter, AB = 13 m.

   Since AB is a diameter, the angle in a semicircle, ∠APB, is 90°. Therefore, ΔAPB is a right-angled triangle.

2. Create the quadratic equation using Pythagoras theorem: In ΔAPB, AP² + BP² = AB² (x + 7)² + x² = 13² (x² + 14x + 49) + x² = 169 2x² + 14x + 49 - 169 = 0 2x² + 14x - 120 = 0 Dividing the entire equation by 2 to simplify, we get: x² + 7x - 60 = 0

3. Check if the situation is possible: To see if it's possible to place the pole, we need to check if the quadratic equation has real roots. We do this by finding the discriminant. Here, a = 1, b = 7, c = -60. D = b² - 4ac D = (7)² - 4(1)(-60) D = 49 + 240 D = 289

   Since D = 289 > 0, the equation has two distinct real roots. Thus, the situation is possible.

4. Solve the equation to find the distances: Using the quadratic formula, x = [-b ± √D] / 2a: x = [-7 ± √289] / (2 * 1) x = [-7 ± 17] / 2

   This gives us two possible values for x:

   • x₁ = (-7 + 17) / 2 = 10 / 2 = 5
   • x₂ = (-7 - 17) / 2 = -24 / 2 = -12

5. Interpret the result: Since x represents a distance, it cannot be negative. Therefore, we discard x = -12. The valid distance is x = 5 metres.

   This means the distance from gate B is BP = x = 5 m. The distance from gate A is AP = x + 7 = 5 + 7 = 12 m.

   Answer: Yes, it is possible to erect the pole. It should be placed at a distance of 5 m from gate B and 12 m from gate A.

Nature of Roots

Chapter 4: Quadratic Equations

Page 5: Nature of Roots

In our previous sections, we learned how to find the roots of a quadratic equation ax² + bx + c = 0 using the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

This formula is powerful because it can solve any quadratic equation. But did you know that a part of this formula can tell us about the roots even before we fully calculate them? It can tell us if the roots are real, if they are distinct, or if they are identical. Let's explore this fascinating concept.

The Discriminant

Look closely at the quadratic formula. The expression inside the square root, b² - 4ac, plays a crucial role. This value is called the discriminant of the quadratic equation.

Definition: For a quadratic equation ax² + bx + c = 0, the discriminant, denoted by D, is given by the expression:

D = b² - 4ac

The word "discriminate" means to recognize a distinction or to differentiate. The discriminant does exactly that—it helps us differentiate between the possible types of roots a quadratic equation can have. The nature of the roots of the equation depends entirely on whether the value of D is positive, zero, or negative.

Let's analyze the three possible cases.

Case 1: Discriminant is Positive (D > 0)

If b² - 4ac > 0, then the value inside the square root, √D, is a positive real number. The quadratic formula then gives us two distinct solutions:

x₁ = (-b + √D) / 2a x₂ = (-b - √D) / 2a

Since D is not zero, adding √D and subtracting √D from -b will result in two different numerators.

Conclusion: If D > 0, the quadratic equation has two distinct real roots.

Case 2: Discriminant is Zero (D = 0)

If b² - 4ac = 0, then the value inside the square root is zero, and √D = √0 = 0. The quadratic formula becomes:

x = [-b ± 0] / 2a

This gives us: x₁ = (-b + 0) / 2a = -b / 2a x₂ = (-b - 0) / 2a = -b / 2a

Both roots are identical. These are also called coincident roots.

Conclusion: If D = 0, the quadratic equation has two equal real roots.

Case 3: Discriminant is Negative (D < 0)

If b² - 4ac < 0, the value inside the square root is negative. In the system of real numbers, we cannot find the square root of a negative number. For example, there is no real number which, when squared, gives -9.

Therefore, we cannot proceed with the calculation to find a real value for x.

Conclusion: If D < 0, the quadratic equation has no real roots.

Summary Table: Nature of Roots

Worked Examples

Let's apply this knowledge to solve some problems.

Example 1 (Based on NCERT Example 7) Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, and hence find the nature of its roots.

Solution: The given equation is 2x² – 4x + 3 = 0.

1. Identify coefficients: Compare the equation with the standard form ax² + bx + c = 0. We have: a = 2, b = -4, c = 3.

2. Calculate the discriminant (D): D = b² - 4ac D = (-4)² - 4(2)(3) D = 16 - 24 D = -8

3. Determine the nature of roots: Since D = -8, which is less than 0 (D < 0), the condition for "no real roots" is met.

Answer: The discriminant is -8. Therefore, the given quadratic equation has no real roots.

Example 2 (Based on NCERT Example 9) Find the discriminant of the equation 3x² – 2x + 1/3 = 0. Find the nature of its roots. If the roots are real, find them.

Solution: The given equation is 3x² – 2x + 1/3 = 0.

1. Identify coefficients: a = 3, b = -2, c = 1/3

2. Calculate the discriminant (D): D = b² - 4ac D = (-2)² - 4(3)(1/3) D = 4 - 4 D = 0

3. Determine the nature of roots: Since D = 0, the quadratic equation has two equal real roots.

4. Find the roots (since they are real): When D = 0, the roots are given by x = -b / 2a. x = -(-2) / (2 * 3) x = 2 / 6 x = 1/3

Answer: The discriminant is 0. The equation has two equal real roots, which are 1/3 and 1/3.

Example 3 (Based on NCERT Exercise 4.3, Q2) Find the value of 'k' for the quadratic equation 2x² + kx + 3 = 0, so that it has two equal roots.

Solution: This is a very common type of question. The key is to use the condition for equal roots.

1. State the condition for equal roots: For a quadratic equation to have two equal roots, its discriminant must be zero. D = b² - 4ac = 0

2. Identify coefficients from the given equation: The equation is 2x² + kx + 3 = 0. a = 2, b = k, c = 3

3. Substitute the coefficients into the discriminant formula and solve for k: (k)² - 4(2)(3) = 0 k² - 24 = 0 k² = 24 k = ±√24

4. Simplify the value of k: √24 = √(4 × 6) = √4 × √6 = 2√6 So, k = ±2√6

Answer: The values of k for which the equation has equal roots are 2√6 and -2√6.

Example 4: Application Problem (Based on NCERT Example 8) Is it possible to erect a pole on the boundary of a circular park of diameter 13 metres such that the difference of its distances from two diametrically opposite gates A and B is 7 metres? If yes, find the distances.

Solution: This problem requires us to first form a quadratic equation and then check if a real solution is possible.

1. Formulate the equation:

   • Let the position of the pole be P on the circle. A and B are ends of a diameter.
   • Let the distance of the pole from gate B be x metres (BP = x).
   • The problem states the difference of distances is 7 m, so AP - BP = 7.
   • This means AP = BP + 7 = x + 7.
   • Since AB is a diameter, the angle in a semi-circle, ∠APB, is 90°.
   • By Pythagoras theorem in ΔAPB: AP² + BP² = AB²
   • (x + 7)² + x² = 13²
   • (x² + 14x + 49) + x² = 169
   • 2x² + 14x + 49 - 169 = 0
   • 2x² + 14x - 120 = 0
   • Dividing by 2, we get: x² + 7x - 60 = 0

2. Check if the situation is possible: To check if it's possible to find a real distance x, we must check if the quadratic equation has real roots. We do this by finding the discriminant.

   • Here, a = 1, b = 7, c = -60.
   • D = b² - 4ac
   • D = (7)² - 4(1)(-60)
   • D = 49 + 240
   • D = 289

3. Interpret the result: Since D = 289 > 0, the equation has two distinct real roots. Therefore, yes, the situation is possible.

4. Find the distances by solving the equation: Using the quadratic formula: x = [-b ± √D] / 2a x = [-7 ± √289] / (2 * 1) x = [-7 ± 17] / 2

   This gives two possible values for x: x₁ = (-7 + 17) / 2 = 10 / 2 = 5 x₂ = (-7 - 17) / 2 = -24 / 2 = -12

   Since x represents a distance, it cannot be negative. We discard x = -12. Therefore, x = 5 metres.

   The distances are:

   • Distance from gate B (BP) = x = 5 m.
   • Distance from gate A (AP) = x + 7 = 5 + 7 = 12 m.

Answer: Yes, it is possible. The pole should be erected at a distance of 5 m from gate B and 12 m from gate A.