Introduction

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

Introduction to Quadratic Equations

Have you ever wondered how architects design a curved dome, or how a satellite dish focuses signals? The shapes they use, called parabolas, are described by a special type of equation. Imagine you're planning a rectangular garden. You have enough soil to cover an area of 150 square meters. You also decide that the length of the garden should be 5 meters longer than its width. How do you find the exact dimensions?

If you let the width be w meters, the length will be w + 5. The area is length × width, which gives us the equation: (w + 5) × w = 150. Expanding this, we get w² + 5w = 150, or w² + 5w - 150 = 0. This is a quadratic equation.

The word "quadratic" comes from the Latin word quadratus, meaning square, because the variable gets squared (like w²). These equations are powerful tools used in physics, engineering, finance, and art to model relationships and solve problems involving curves and optimization. In this chapter, we will learn how to identify, formulate, and, most importantly, solve these second-degree polynomial equations.

Definitions & Formulas

Understanding the language of quadratic equations is the first step. The standard form is the universally accepted format to write and analyze these equations.

{{KEY: type=concept | title=The Non-Negotiable Rule | text=For an equation to be quadratic, the coefficient of the x² term, 'a', must be non-zero (a ≠ 0). If a = 0, the x² term vanishes, and the equation becomes a linear equation (bx + c = 0), not a quadratic one.}}

Logic: Converting to Standard Form

Before you can solve a quadratic equation, you must be able to write it in its standard form: ax² + bx + c = 0. This process involves simple algebraic manipulation. Let's break down the logic with an example equation: (x - 2)² + 1 = 2x - 3.

1. Eliminate Brackets and Powers: First, expand any squared terms or brackets. In our example, we expand (x - 2)² using the identity (a - b)² = a² - 2ab + b².

   x² - 4x + 4 + 1 = 2x - 3
   

2. Simplify Each Side: Combine any constant terms or like terms on each side of the equation. Here, we can combine 4 and 1.

   x² - 4x + 5 = 2x - 3
   

3. Move All Terms to One Side: The goal is to have 0 on one side of the equation, typically the right side. To do this, subtract 2x and add 3 to both sides.

   x² - 4x + 5 - 2x + 3 = 0
   

4. Combine Like Terms: Now, group and combine the x² terms, the x terms, and the constant terms.

   x² + (-4x - 2x) + (5 + 3) = 0
   

5. Write the Final Standard Form: Perform the final arithmetic to get the clean standard form.

   x² - 6x + 8 = 0
   

6. Identify Coefficients: Now that it's in the form ax² + bx + c = 0, we can easily identify the coefficients. Here, a = 1, b = -6, and c = 8. Since a ≠ 0, this is indeed a quadratic equation.

Solved Examples

Let's practice identifying and formulating quadratic equations.

Example 1: Basic Identification (Easy)

Given: The equation x(2x + 3) = x² + 1.

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

Solution:

1. Start by expanding the expression on the left-hand side (LHS).

   2x² + 3x = x² + 1
   

2. Bring all terms from the right-hand side (RHS) to the LHS to set the equation equal to zero.

   2x² + 3x - x² - 1 = 0
   

3. Combine the like terms (2x² and -x²).

   x² + 3x - 1 = 0
   

4. Compare the resulting equation with the standard form ax² + bx + c = 0. We can see that a = 1, b = 3, and c = -1.

5. The highest power of the variable x is 2, and the coefficient of x² (a = 1) is not zero. Therefore, the equation is a quadratic equation.

Final Answer:

Yes, the equation x(2x + 3) = x² + 1 is a quadratic equation.

Example 2: Simplification with Cubes (Medium)

Given: The equation (x + 2)³ = 2x(x² - 1).

To Find: Determine if this equation is quadratic.

Solution:

1. Expand the expressions on both sides. Use the identity (a + b)³ = a³ + b³ + 3ab(a + b) for the LHS.

   x³ + 2³ + 3(x)(2)(x + 2) = 2x³ - 2x
   

2. Simplify the expanded LHS.

   x³ + 8 + 6x(x + 2) = 2x³ - 2x
   

   x³ + 8 + 6x² + 12x = 2x³ - 2x
   

3. Move all terms to one side to set the equation to zero. Let's move them to the RHS to keep the leading term positive.

   0 = 2x³ - 2x - (x³ + 6x² + 12x + 8)
   

4. Distribute the negative sign and combine like terms.

   0 = 2x³ - 2x - x³ - 6x² - 12x - 8
   

   0 = (2x³ - x³) - 6x² + (-2x - 12x) - 8
   

5. Simplify to get the final form.

   x³ - 6x² - 14x - 8 = 0
   

6. Check the highest power of the variable x. The highest power is 3. An equation with degree 3 is a cubic equation, not a quadratic one.

Final Answer:

No, the equation (x + 2)³ = 2x(x² - 1) is not a quadratic equation; it is a cubic equation.

Example 3: Formulating from a Word Problem (Hard)

Given: The product of two consecutive positive integers is 306.

To Find: Formulate the quadratic equation to find the integers.

Solution:

1. Let the first positive integer be x.

2. Since the integers are consecutive, the next integer will be x + 1.

3. The problem states that their product is 306. We can write this as an equation.

   x × (x + 1) = 306
   

4. Expand the left side of the equation.

   x² + x = 306
   

5. Convert the equation to the standard form ax² + bx + c = 0 by subtracting 306 from both sides.

   x² + x - 306 = 0
   

6. This equation is in the standard quadratic form, where a = 1, b = 1, and c = -306. This equation can be solved to find the value of x.

Final Answer:

The required quadratic equation is x² + x - 306 = 0.

Example 4: Real-World Application (Tricky)

Given: A rectangular hall has a floor with a perimeter of 92 meters and an area of 480 square meters.

To Find: Formulate the quadratic equation for the length and breadth of the hall.

Solution:

1. Let the length of the hall be l and the breadth be b.

2. The formula for the perimeter of a rectangle is 2(l + b). We are given that the perimeter is 92 m.

   2(l + b) = 92
   

   l + b = 46
   

3. From the perimeter equation, we can express one variable in terms of the other. Let's express l.

   l = 46 - b
   

4. The formula for the area of a rectangle is l × b. We are given that the area is 480 m².

   l × b = 480
   

5. Substitute the expression for l from step 3 into the area equation. This will create an equation with a single variable, b.

   (46 - b) × b = 480
   

6. Expand the equation.

   46b - b² = 480
   

7. Rearrange the terms to get the standard quadratic form ax² + bx + c = 0. It's conventional to make the b² term positive.

   0 = b² - 46b + 480
   

Final Answer:

The required quadratic equation in terms of breadth 'b' is b² - 46b + 480 = 0.

Tips & Tricks

Mastering these small techniques can save you time and prevent errors.

Common Mistakes

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

Brain-Teaser Questions

Test your understanding with these slightly challenging questions.

1. For what value of m does the equation (m² - 4)x² + 5x + 9 = 0 cease to be a quadratic equation?

   💡 Answer: An equation is not quadratic if the coefficient of x² is zero. So, we set m² - 4 = 0. This gives m² = 4, which means m = 2 or m = -2. For these two values of m, the equation becomes linear.

2. A student simplifies the equation x + 3/x = x². They multiply the entire equation by x to get x² + 3 = x³. Is the original equation a quadratic equation?

   💡 Answer: No. A quadratic equation must be a polynomial equation of degree 2. The original equation x + 3/x = x² has a term 3/x (or 3x⁻¹), which has a negative exponent. This means the original equation is not a polynomial, and therefore cannot be a quadratic equation.

3. If a, b, and c are integers in the equation ax² + bx + c = 0, can a quadratic equation have fractional roots (solutions)? Give an example.

   💡 Answer: Yes, absolutely. The coefficients being integers does not mean the roots must be integers. Consider the equation 2x² + 5x - 3 = 0. Here a=2, b=5, c=-3 are all integers. This equation can be factored as (2x - 1)(x + 3) = 0. The solutions (roots) are x = 1/2 and x = -3. One of the roots is a fraction.

Mini Cheatsheet

A quick summary of today's key concepts for last-minute revision.

Quadratic Equations

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

Page 2: Solving Quadratic Equations & Nature of Roots

Welcome back! In the previous section, we learned to identify quadratic equations. Now, let's get to the most exciting part: finding their solutions, also known as roots. Think of it like a detective story where we need to find the value(s) of x that make the equation true.

Imagine an architect designing an arched bridge. The shape of the arch is a parabola, which can be described by a quadratic equation. To find the points where the bridge touches the ground, the architect needs to solve the equation for when the height is zero. These points are the roots of the equation! We will explore three powerful methods to find these roots: Factorization, Completing the Square, and the universal Quadratic Formula.

Definitions & Formulas

Before we jump into solving, let's get our key terms and formulas in one place. The discriminant is a special value that tells us about the nature of the roots without actually solving for them.

Derivation: The Quadratic Formula

Ever wondered where the mighty quadratic formula comes from? It isn't magic! It's derived from the standard form of the equation using a technique called "Completing the Square". Let's derive it step-by-step.

Starting Point: The standard form of a quadratic equation.

ax² + bx + c = 0

1. Isolate the x² and x terms. We move the constant c to the right side of the equation.

   ax² + bx = -c
   

2. Make the coefficient of x² equal to 1. We achieve this by dividing the entire equation by a.

   x² + (b/a)x = -c/a
   

3. Complete the square. This is the key step. We add the square of half the coefficient of x to both sides. The coefficient of x is b/a. Half of it is b/2a. Its square is (b/2a)².

   x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
   

4. Factor the left side. The left side is now a perfect square trinomial, which can be factored as (x + b/2a)². Let's also simplify the right side.

   (x + b/2a)² = -c/a + b²/(4a²)
   

   To combine the terms on the right, we find a common denominator (4a²).

   (x + b/2a)² = (-4ac + b²)/(4a²)
   

5. Take the square root of both sides. This helps us to start isolating x. Remember to include ± on the right side.

   x + b/2a = ±√(b² - 4ac) / √(4a²)
   

   Simplifying the denominator gives:

   x + b/2a = ±√(b² - 4ac) / 2a
   

6. Solve for x. Finally, we isolate x by moving b/2a to the right side.

   x = -b/2a ± √(b² - 4ac) / 2a
   

   Combining the terms over the common denominator 2a gives us the famous Quadratic Formula.

   x = (-b ± √(b² - 4ac)) / 2a
   

{{KEY: type=concept | title=The Discriminant and Nature of Roots | text=The value inside the square root in the quadratic formula, D = b² - 4ac, is the Discriminant. It tells you exactly how many real solutions the equation has. If D > 0, you get two distinct real roots. If D = 0, you get exactly one real root (or two equal real roots). If D < 0, you get no real roots (the solutions are complex numbers, which you'll study in Class 11).}}

Solved Examples

Let's apply these methods to some problems, starting from easy and moving to more challenging ones.

Example 1: Solving by Factorization (Easy)

Given: The equation x² - 3x - 10 = 0.

To Find: The roots of the equation.

Solution:

1. We need to find two numbers that multiply to ac (1 × -10 = -10) and add up to b (-3). These numbers are -5 and 2.

2. Split the middle term -3x into -5x + 2x.

   x² - 5x + 2x - 10 = 0
   

3. Factor by grouping. Take x common from the first two terms and 2 common from the last two.

   x(x - 5) + 2(x - 5) = 0
   

4. Take the common factor (x - 5) out.

   (x - 5)(x + 2) = 0
   

5. Set each factor to zero to find the roots.

   x - 5 = 0  OR  x + 2 = 0
   

   x = 5  OR  x = -2
   

Final Answer: The roots are 5 and -2.

Example 2: Solving by Completing the Square (Medium)

Given: The equation 2x² - 7x + 3 = 0.

To Find: The roots of the equation.

Solution:

1. Divide the entire equation by the coefficient of x², which is 2.

   x² - (7/2)x + 3/2 = 0
   

2. Move the constant term to the right side.

   x² - (7/2)x = -3/2
   

3. Find half of the coefficient of x and square it. Half of -7/2 is -7/4. Squaring it gives (-7/4)² = 49/16. Add this to both sides.

   x² - (7/2)x + 49/16 = -3/2 + 49/16
   

4. Factor the left side as a perfect square and simplify the right side.

   (x - 7/4)² = -24/16 + 49/16
   

   (x - 7/4)² = 25/16
   

5. Take the square root of both sides.

   x - 7/4 = ±√(25/16)
   

   x - 7/4 = ±5/4
   

6. Solve for x by considering both the + and - cases.

   Case 1: x - 7/4 = 5/4 → x = 5/4 + 7/4 = 12/4 = 3 Case 2: x - 7/4 = -5/4 → x = -5/4 + 7/4 = 2/4 = 1/2

Final Answer: The roots are 3 and 1/2.

Example 3: Using the Quadratic Formula (Hard)

Given: The equation 3x² + 5x - 4 = 0.

To Find: The roots of the equation.

Solution:

1. Compare the given equation with the standard form ax² + bx + c = 0 to identify the coefficients.

   Here, a = 3, b = 5, c = -4.

2. Write down the quadratic formula.

   x = (-b ± √(b² - 4ac)) / 2a
   

3. Substitute the values of a, b, and c into the formula.

   x = (-(5) ± √((5)² - 4 × 3 × (-4))) / (2 × 3)
   

4. Simplify the expression inside the square root (the discriminant).

   x = (-5 ± √(25 - (-48))) / 6
   

   x = (-5 ± √(25 + 48)) / 6
   

   x = (-5 ± √73) / 6
   

5. Since 73 is a prime number, its square root cannot be simplified further. So we write the two distinct roots.

   x = (-5 + √73) / 6 and x = (-5 - √73) / 6

Final Answer: The roots are (-5 + √73) / 6 and (-5 - √73) / 6.

Example 4: Finding a Variable using Nature of Roots (Tricky)

Given: The quadratic equation kx² - 8x + 4 = 0 has real and equal roots.

To Find: The value of k.

Solution:

1. The condition for a quadratic equation to have real and equal roots is that its discriminant (D) must be zero.

   D = b² - 4ac = 0
   

2. Identify the coefficients a, b, and c from the given equation kx² - 8x + 4 = 0.

   Here, a = k, b = -8, c = 4.

3. Substitute these values into the discriminant formula and set it to zero.

   (-8)² - 4(k)(4) = 0
   

4. Solve the resulting equation for k.

   64 - 16k = 0
   

   64 = 16k
   

   k = 64 / 16
   

   k = 4
   

Final Answer: The value of k is 4.

Tips & Tricks

Work smarter, not just harder! Here are some shortcuts to speed up your problem-solving.

Common Mistakes

Many students lose marks because of small, avoidable errors. Here are the most common pitfalls.

Brain-Teaser Questions

Ready for a challenge? These questions require you to apply the concepts in a slightly different way.

1. If α and β are the roots of the equation 2x² - 6x + 3 = 0, what is the value of α² + β² without finding the roots?

   💡 Answer: We know α+β = -b/a = -(-6)/2 = 3, and αβ = c/a = 3/2. Also, (α+β)² = α² + β² + 2αβ. So, α² + β² = (α+β)² - 2αβ. Substituting the values: (3)² - 2(3/2) = 9 - 3 = 6.

2. For what positive value of p, does the equation x² + px + 64 = 0 and x² - 8x + p = 0 both have real roots?

   💡 Answer: For the first equation, D₁ ≥ 0 → p² - 4(1)(64) ≥ 0 → p² - 256 ≥ 0 → p² ≥ 256. This means p ≥ 16 or p ≤ -16. For the second equation, D₂ ≥ 0 → (-8)² - 4(1)(p) ≥ 0 → 64 - 4p ≥ 0 → 64 ≥ 4p → 16 ≥ p. We need a positive p that satisfies both p ≥ 16 and p ≤ 16. The only value that satisfies both is p = 16.

3. If the roots of the equation (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 are equal, prove that either a = 0 or a³ + b³ + c³ = 3abc.

   💡 Answer: This is a very tough one! If roots are equal, D=0. [-2(a²-bc)]² - 4(c²-ab)(b²-ac) = 0 4(a⁴ + b²c² - 2a²bc) - 4(b²c² - ac³ - ab³ + a²bc) = 0 Dividing by 4 and simplifying gives: a⁴ + b²c² - 2a²bc - b²c² + ac³ + ab³ - a²bc = 0 a⁴ + ab³ + ac³ - 3a²bc = 0 Taking a common: a(a³ + b³ + c³ - 3abc) = 0. This implies that either a = 0 or a³ + b³ + c³ - 3abc = 0, which means a³ + b³ + c³ = 3abc.

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for last-minute revision!

Solution of a Quadratic Equation by Factorisation — Part 1

Chapter 4: Quadratic Equations | Page 3 of 5

Solution of a Quadratic Equation by Factorisation — Part 1

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Concept Introduction

Imagine you're designing a rectangular community garden. The local council has specified that the length of the garden must be 5 meters more than its width, and the total area must be exactly 84 square meters. How would you find the exact dimensions?

If you let the width be x meters, the length becomes x + 5 meters. The area is length × width, which gives us the equation: x(x + 5) = 84. Expanding this, we get x² + 5x - 84 = 0. This is a quadratic equation.

To find the value of x (the width), we need to solve this equation. One of the most intuitive methods to do this is factorisation. It's like breaking a code. We try to reverse the multiplication process to find two simpler expressions that, when multiplied, give us our original quadratic equation. By solving these simpler expressions, we can find the dimensions of our garden.

Definitions & Formulas

Understanding the terminology is the first step to mastering the method.

The Logic Behind Factorisation

The "Splitting the Middle Term" method is a systematic way to factorize a quadratic expression. It's based on reversing the FOIL (First, Outer, Inner, Last) method you use for multiplying two binomials.

Let's break down the logic:

1. Start with the standard form. Our goal is to solve the equation:

   ax² + bx + c = 0
   

2. Identify the target. We need to find two numbers, let's call them p and q, that will help us break down the middle term bx.

3. Set the conditions. These two numbers, p and q, must satisfy two specific conditions:

   • Their sum must be equal to the middle coefficient, b.

     p + q = b
     

   • Their product must be equal to the product of the first and last coefficients, a and c.

     p × q = a × c
     

4. Split the middle term. We rewrite the original equation by replacing bx with px + qx.

   ax² + px + qx + c = 0
   

5. Factor by grouping. Now, we group the first two terms and the last two terms and factor out the greatest common factor from each pair.

   (ax² + px) + (qx + c) = 0
   

   This will lead to a common binomial factor, which can then be factored out. For example, x(ax + p) + y(ax + p) = 0, which becomes (x+y)(ax+p) = 0.

6. Apply the Zero Product Property. Once we have the equation in its factored form, (Factor 1) × (Factor 2) = 0, we can set each factor equal to zero and solve for x. This gives us the two roots of the quadratic equation.

{{KEY: type=concept | title=The Core Idea of Factorisation | text=The method revolves around finding two numbers whose sum is 'b' and whose product is 'ac'. This allows us to split the middle term, group the terms, and extract common factors to find the roots.}}

Solved Examples

Let's apply this method to some problems, starting from easy and moving to more challenging ones.

Example 1: Basic Factorisation (a = 1)

Given: The quadratic equation x² + 7x + 12 = 0.

To Find: The roots of the equation.

Solution:

1. Compare the equation with ax² + bx + c = 0. We have a = 1, b = 7, and c = 12.

2. We need to find two numbers, p and q, such that p + q = 7 and p × q = a × c = 1 × 12 = 12.

3. Let's list the factors of 12: (1, 12), (2, 6), (3, 4). The pair (3, 4) adds up to 7. So, p = 3 and q = 4.

4. Split the middle term 7x into 3x + 4x.

   x² + 3x + 4x + 12 = 0
   

5. Factor by grouping the terms.

   (x² + 3x) + (4x + 12) = 0
   

6. Factor out the common term from each group.

   x(x + 3) + 4(x + 3) = 0
   

7. Factor out the common binomial (x + 3).

   (x + 3)(x + 4) = 0
   

8. Apply the Zero Product Property. Either x + 3 = 0 or x + 4 = 0.

   x = -3   or   x = -4
   

Final Answer:

The roots are -3 and -4.

Example 2: Medium Difficulty (a ≠ 1)

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

To Find: The roots of the equation.

Solution:

1. Here, a = 6, b = -1, and c = -2.

2. We need two numbers p and q such that p + q = -1 and p × q = a × c = 6 × (-2) = -12.

3. Since the product p × q is negative, the numbers must have opposite signs. Since the sum p + q is negative, the larger number must be negative. Factors of 12: (1, 12), (2, 6), (3, 4). The pair that gives a difference of 1 is (3, 4). To get a sum of -1, we need p = 3 and q = -4.

4. Split the middle term -x into 3x - 4x.

   6x² + 3x - 4x - 2 = 0
   

5. Factor by grouping.

   (6x² + 3x) - (4x + 2) = 0
   

   Note the sign change in the second bracket because we factored out a negative.

6. Factor out the common term from each group.

   3x(2x + 1) - 2(2x + 1) = 0
   

7. Factor out the common binomial (2x + 1).

   (2x + 1)(3x - 2) = 0
   

8. Apply the Zero Product Property. Either 2x + 1 = 0 or 3x - 2 = 0.

   If 2x + 1 = 0, then 2x = -1, so x = -1/2. If 3x - 2 = 0, then 3x = 2, so x = 2/3.

Final Answer:

The roots are -1/2 and 2/3.

Example 3: Hard (Roots in Coefficients)

Given: The quadratic equation √2x² + 7x + 5√2 = 0.

To Find: The roots of the equation.

Solution:

1. This looks intimidating, but the process is the same. Here, a = √2, b = 7, and c = 5√2.

2. First, find the product a × c.

   a × c = (√2) × (5√2) = 5 × (√2 × √2) = 5 × 2 = 10
   

3. Now, we need two numbers p and q such that p + q = 7 and p × q = 10. By inspection, the numbers are 2 and 5. So, p = 2 and q = 5.

4. Split the middle term 7x into 2x + 5x.

   √2x² + 2x + 5x + 5√2 = 0
   

5. Factor by grouping. This is the crucial step. We can write 2 as √2 × √2 to find a common factor.

   (√2x² + √2⋅√2x) + (5x + 5√2) = 0
   

6. Factor out the common term from each group.

   √2x(x + √2) + 5(x + √2) = 0
   

7. Factor out the common binomial (x + √2).

   (x + √2)(√2x + 5) = 0
   

8. Apply the Zero Product Property. Either x + √2 = 0 or √2x + 5 = 0.

   If x + √2 = 0, then x = -√2. If √2x + 5 = 0, then √2x = -5, so x = -5/√2.

Final Answer:

The roots are -√2 and -5/√2.

Example 4: Tricky (Algebraic Coefficients)

Given: The equation 4x² - 4a²x + (a⁴ - b⁴) = 0.

To Find: The roots of the equation in terms of a and b.

Solution:

1. Here, the coefficients are algebraic expressions. A = 4, B = -4a², and C = a⁴ - b⁴.

2. First, find the product A × C.

   A × C = 4 × (a⁴ - b⁴)
   

   We can factor a⁴ - b⁴ using the difference of squares identity: a⁴ - b⁴ = (a² - b²)(a² + b²). So, A × C = 4(a² - b²)(a² + b²).

3. Now, we need two numbers p and q such that their sum is B = -4a² and their product is A × C. This is the trickiest part. Let's try splitting the product A × C into two factors that might add up to -4a². Let's try p = -2(a² - b²) and q = -2(a² + b²).

4. Let's check their sum:

   p + q = -2(a² - b²) - 2(a² + b²)
         = -2a² + 2b² - 2a² - 2b²
         = -4a²
   

   This matches B. So we have found our p and q.

5. Split the middle term -4a²x.

   4x² - 2(a² - b²)x - 2(a² + b²)x + (a² - b²)(a² + b²) = 0
   

6. Factor by grouping.

   [4x² - 2(a² - b²)x] - [2(a² + b²)x - (a² - b²)(a² + b²)] = 0
   

   Factor out 2x from the first group and (a² + b²) from the second.

   2x[2x - (a² - b²)] - (a² + b²)[2x - (a² - b²)] = 0
   

7. Factor out the common binomial [2x - (a² - b²)].

   [2x - (a² - b²)] [2x - (a² + b²)] = 0
   

8. Apply the Zero Product Property. Either 2x - (a² - b²) = 0 or 2x - (a² + b²) = 0.

   If 2x - (a² - b²) = 0, then 2x = a² - b², so x = (a² - b²)/2. If 2x - (a² + b²) = 0, then 2x = a² + b², so x = (a² + b²)/2.

Final Answer:

The roots are (a² - b²)/2 and (a² + b²)/2.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Solve for x: abx² + (b² - ac)x - bc = 0

   💡 Answer: Split the middle term: abx² + b²x - acx - bc = 0. Group: (abx² + b²x) - (acx + bc) = 0. Factor: bx(ax + b) - c(ax + b) = 0. Factor again: (bx - c)(ax + b) = 0. Roots: x = c/b and x = -b/a.

2. The sum of a number and its positive square root is 6. Find the number.

   💡 Answer: Let the number be x. The equation is x + √x = 6. Let y = √x. Then y² = x. The equation becomes y² + y = 6, or y² + y - 6 = 0. Factorising: (y + 3)(y - 2) = 0. So y = -3 or y = 2. Since y is a positive square root, y must be 2. As x = y², the number is x = 2² = 4.

3. Solve the equation: (x + 3)² - 4(x + 3) - 5 = 0.

   💡 Answer: This looks complex, but use substitution. Let y = (x + 3). The equation becomes a simple quadratic: y² - 4y - 5 = 0. Factorise it: (y - 5)(y + 1) = 0. So, y = 5 or y = -1. Now substitute back. Case 1: x + 3 = 5 → x = 2. Case 2: x + 3 = -1 → x = -4. The roots are 2 and -4.

Mini Cheatsheet

Solution of a Quadratic Equation by Factorisation — Part 2

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

Page 4: Solution of a Quadratic Equation by Factorisation — Part 2

Welcome back! In the previous section, we were introduced to the idea of solving quadratic equations by factorisation. Now, we'll master a powerful and systematic technique called splitting the middle term. This method works for almost any factorisable quadratic equation you'll encounter in your exams and is a foundational skill in algebra.

Imagine an architect designing a rectangular community hall. She knows the total required area is 300 square meters. She also decides that the length of the hall must be 5 meters more than twice its width. How can she find the exact dimensions? This problem translates directly into a quadratic equation: if the width is w, the length is 2w + 5, and the area is w(2w + 5) = 300, which simplifies to 2w² + 5w - 300 = 0. To find the width w, we need to solve this equation. Let's learn the method that makes solving such problems straightforward.

Definitions & Formulas

Let's revisit the key terms that form the foundation of our method.

{{KEY: type=concept | title=Splitting the Middle Term | text=This is the core technique of factorisation. We rewrite the middle term, bx, as a sum or difference of two new terms. This allows us to group terms and find common factors, ultimately expressing the quadratic as a product of two linear factors.}}

The Logic of Splitting the Middle Term

Why does this method work? It's a clever reversal of the multiplication process. Let's break down the logic.

1. Start with the standard quadratic expression ax² + bx + c. Our goal is to convert it into the form (px + q)(rx + s).

2. If we multiply out (px + q)(rx + s), we get:

   (pr)x² + (ps + qr)x + qs
   

3. Now, compare this expanded form with our original expression ax² + bx + c.

   • a = pr
   • b = ps + qr
   • c = qs

4. Let's look at the product of the coefficient of x² and the constant term, ac:

   ac = (pr) × (qs) = pqrs
   

5. Notice that the two terms we "split" the middle term into are ps and qr. Their sum is b. What is their product?

   (ps) × (qr) = pqrs
   

6. This is the magic! The product of the two new terms (ps and qr) is the same as the product ac. This is why the method works. We need to find two numbers that:

   • Add up to b (the middle coefficient)
   • Multiply to ac (the product of the first and last coefficients)

Solved Examples

Let's apply this logic to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Basic Factorisation (a=1)

Given: The quadratic equation x² - 3x - 10 = 0.

To Find: The roots of the equation.

Solution:

1. Compare the equation with ax² + bx + c = 0. We have a = 1, b = -3, and c = -10.

2. We need to find two numbers whose sum is b = -3 and product is ac = 1 × (-10) = -10. Let's think of factors of -10: (1, -10), (-1, 10), (2, -5), (-2, 5). The pair 2 and -5 works because 2 + (-5) = -3.

3. Split the middle term -3x into +2x - 5x.

   x² + 2x - 5x - 10 = 0
   

4. Group the terms and factor out the common factors from each group.

   (x² + 2x) - (5x + 10) = 0
   

   x(x + 2) - 5(x + 2) = 0
   

5. Now, (x + 2) is a common factor.

   (x - 5)(x + 2) = 0
   

6. Apply the Zero Product Property. Either (x - 5) = 0 or (x + 2) = 0.

   x = 5  or  x = -2
   

Final Answer:

The roots are 5 and -2.

Example 2: Medium Difficulty (a > 1)

Given: The quadratic equation 3x² - x - 4 = 0.

To Find: The roots of the equation.

Solution:

1. Here, a = 3, b = -1, and c = -4.

2. We need two numbers whose sum is b = -1 and product is ac = 3 × (-4) = -12. The factor pairs of -12 are (1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4). The pair 3 and -4 fits our requirement as 3 + (-4) = -1.

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

   3x² + 3x - 4x - 4 = 0
   

4. Group the terms and factor out common factors.

   (3x² + 3x) - (4x + 4) = 0
   

   3x(x + 1) - 4(x + 1) = 0
   

5. Take (x + 1) as the common factor.

   (3x - 4)(x + 1) = 0
   

6. Using the Zero Product Property:

   • 3x - 4 = 0 → 3x = 4 → x = 4/3
   • x + 1 = 0 → x = -1

Final Answer:

The roots are 4/3 and -1.

Example 3: Hard (Coefficients with Square Roots)

Given: The quadratic equation √2x² + 7x + 5√2 = 0.

To Find: The roots of the equation.

Solution:

1. This looks intimidating, but the method is the same. Here, a = √2, b = 7, and c = 5√2.

2. First, calculate the product ac.

   ac = (√2) × (5√2) = 5 × (√2 × √2) = 5 × 2 = 10
   

3. We need two numbers that sum to b = 7 and multiply to ac = 10. The numbers are clearly 5 and 2.

4. Split the middle term 7x into 5x + 2x.

   √2x² + 5x + 2x + 5√2 = 0
   

   A small tip: It's often easier to group the term with √2 next to the other term with √2. Let's rearrange.

   √2x² + 2x + 5x + 5√2 = 0
   

5. Group and factor. Remember that 2 can be written as √2 × √2.

   (√2x² + 2x) + (5x + 5√2) = 0
   

   √2x(x + √2) + 5(x + √2) = 0
   

6. Now, (x + √2) is the common factor.

   (√2x + 5)(x + √2) = 0
   

7. Apply the Zero Product Property:

   • √2x + 5 = 0 → √2x = -5 → x = -5/√2
   • x + √2 = 0 → x = -√2

Final Answer:

The roots are -5/√2 and -√2.

Example 4: Tricky (Requires Simplification First)

Given: The equation 4/x - 3 = 5/(2x + 3), where x ≠ 0 and x ≠ -3/2.

To Find: The roots of the equation.

Solution:

1. First, we need to convert this equation into the standard quadratic form ax² + bx + c = 0.

2. Simplify the left side by taking a common denominator.

   (4 - 3x)/x = 5/(2x + 3)
   

3. Cross-multiply to eliminate the fractions.

   (4 - 3x)(2x + 3) = 5x
   

4. Expand the expression on the left side.

   4(2x + 3) - 3x(2x + 3) = 5x
   

   8x + 12 - 6x² - 9x = 5x
   

5. Combine like terms and rearrange to the standard form.

   -6x² - x + 12 = 5x
   

   -6x² - x - 5x + 12 = 0
   

   -6x² - 6x + 12 = 0
   

6. To make factorization easier, divide the entire equation by -6.

   x² + x - 2 = 0
   

7. Now, we factor this simple quadratic. We need two numbers that sum to 1 and multiply to -2. The numbers are +2 and -1.

   x² + 2x - x - 2 = 0
   

   x(x + 2) - 1(x + 2) = 0
   

   (x - 1)(x + 2) = 0
   

8. Using the Zero Product Property: x - 1 = 0 or x + 2 = 0.

   x = 1  or  x = -2
   

Final Answer:

The roots are 1 and -2.

Tips & Tricks

Here are a few shortcuts to make your factorization process faster and more accurate.

Common Mistakes

Many students make small, avoidable errors. Here’s how to spot and fix them.

Brain-Teaser Questions

Test your understanding with these challenging problems!

1. Solve for x by factorisation: abx² + (b² - ac)x - bc = 0

   💡 Answer: The middle term is already "split" for you! Group the terms: (abx² + b²x) - (acx + bc) = 0. Factor out common terms: bx(ax + b) - c(ax + b) = 0. This gives (bx - c)(ax + b) = 0. The roots are x = c/b and x = -b/a.

2. Find the roots of the equation: x² - (√3 + 1)x + √3 = 0

   💡 Answer: Here a=1, b = -(√3 + 1), and c = √3. We need two numbers that sum to -(√3 + 1) and multiply to √3. The numbers are -√3 and -1. Split the middle term: x² - √3x - 1x + √3 = 0. Group and factor: x(x - √3) - 1(x - √3) = 0. This gives (x - 1)(x - √3) = 0. The roots are x = 1 and x = √3.

3. A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits are reversed. Find the number.

   💡 Answer: Let the digits be x (tens place) and y (units place). The number is 10x + y. Given: x × y = 14, so y = 14/x. Also given: (10x + y) + 45 = 10y + x. Simplify: 9x - 9y + 45 = 0 → x - y + 5 = 0. Substitute y = 14/x: x - (14/x) + 5 = 0. Multiply by x: x² - 14 + 5x = 0 → x² + 5x - 14 = 0. Factorise: (x + 7)(x - 2) = 0. Since a digit cannot be negative, x = 2. Then y = 14/2 = 7. The number is 27. (Check: 27 + 45 = 72, which is 27 reversed).

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for your last-minute revision!

Nature of Roots

{{FORMULA: expr=D = b² − 4ac | symbols=D:Discriminant, a:quadratic coefficient, b:linear coefficient, c:constant term}}

Page 5: Nature of Roots

Welcome to the final, and perhaps most insightful, part of our journey through Quadratic Equations! So far, we've learned how to find the roots (or solutions) of these equations. But what if we could predict the type of solutions we'll get without actually solving the equation?

Imagine a cricket match. A batsman hits the ball high into the air. The path of the ball is a perfect parabola, which can be described by a quadratic equation. Now, suppose we want to know if the ball will reach a specific height, say, the top of the stadium lights. We could set up an equation. This equation might have two solutions (the ball reaches that height on its way up and on its way down), one solution (it just grazes that height at its peak), or no real solution at all (it never reaches that height). Understanding the "nature of roots" allows us to answer this question quickly, by analyzing the equation itself, without calculating the exact time or position. It's like having a crystal ball for our quadratic equations!

Definitions & Formulas

The ability to predict the nature of roots hinges on one single, powerful value. This value is called the discriminant. It discriminates (or tells the difference) between the possible types of roots.

{{KEY: type=concept | title=The Power of the Discriminant | text=The value of D (b² - 4ac) is the key. A positive D means two different real roots. A zero D means two identical real roots. A negative D means no real roots exist.}}

The Logic Behind the Discriminant

Why does this one value, D = b² - 4ac, hold so much power? The answer lies directly within the Quadratic Formula we learned earlier.

1. Recall the Quadratic Formula The formula to find the roots x of the equation ax² + bx + c = 0 is:

   x = (-b ± √(b² - 4ac)) / 2a
   

2. Isolate the Deciding Factor Look closely at the part under the square root: √(b² - 4ac). The behaviour of a square root is very specific. You can only take the square root of non-negative numbers in the real number system. This single component is what changes the outcome. Let's call this part the discriminant, D.

   D = b² - 4ac
   

   So the formula becomes:

   x = (-b ± √D) / 2a
   

3. Case 1: The Discriminant is Positive (D > 0) If D is a positive number (like 9, 25, or 10), then √D is a real, positive number (like 3, 5, or √10). The ± sign in the formula becomes active. This gives us two distinct, real roots:

   x₁ = (-b + √D) / 2a
   

   x₂ = (-b - √D) / 2a
   

   Conclusion: Two distinct real roots.

4. Case 2: The Discriminant is Zero (D = 0) If D is exactly zero, then √D is also zero. The ± part of the formula effectively vanishes because adding or subtracting zero doesn't change the value.

   x = (-b ± √0) / 2a
   

   This simplifies to:

   x = -b / 2a
   

   We get only one value for the root. However, since a quadratic equation always has two roots, we say that it has two real and equal roots. Conclusion: Two equal real roots.

5. Case 3: The Discriminant is Negative (D < 0) If D is a negative number (like -4 or -11), we encounter a problem. In the system of real numbers (which is what we use in Class 10), the square root of a negative number is not defined. You cannot find a real number that, when squared, gives a negative result. Since √D is not a real number, the entire formula for x does not produce any real solutions. Conclusion: No real roots exist.

Solved Examples

Let's apply this concept to see how it works in practice.

Example 1: Basic Check (Easy)

Given: The quadratic equation 2x² - 6x + 5 = 0.

To Find: The nature of its roots.

Solution:

1. Compare the equation with the standard form ax² + bx + c = 0 to identify the coefficients.

   a = 2, b = -6, c = 5
   

2. Calculate the discriminant, D = b² - 4ac. Be careful with the signs.

   D = (-6)² - 4(2)(5)
   

3. Simplify the expression.

   D = 36 - 40
   

   D = -4
   

4. Analyze the value of D. Since D is negative (-4 < 0), the equation has no real roots.

Final Answer: Since D = -4 < 0, the equation has no real roots.

Example 2: Finding a Missing Value (Medium)

Given: The quadratic equation kx² - 8x + k = 0 has real and equal roots.

To Find: The possible values of k.

Solution:

1. Identify the coefficients from the equation.

   a = k, b = -8, c = k
   

2. The problem states that the roots are real and equal. This gives us a specific condition for the discriminant.

   D = 0
   

3. Set up the discriminant formula and equate it to zero.

   b² - 4ac = 0
   

4. Substitute the coefficient values into the equation.

   (-8)² - 4(k)(k) = 0
   

5. Solve the resulting equation for k.

   64 - 4k² = 0
   

   64 = 4k²
   

   k² = 64 / 4
   

   k² = 16
   

6. Take the square root to find the values of k.

   k = ±√16
   

   k = 4  or  k = -4
   

Final Answer: The possible values of k are 4 and -4.

Example 3: Working with Conditions (Hard)

Given: The quadratic equation (p+1)x² - 6(p+1)x + 3(p+9) = 0, where p ≠ -1, has equal roots.

To Find: The value of p.

Solution:

1. This looks complicated, but the process is the same. First, identify a, b, and c.

   a = (p+1)
   b = -6(p+1)
   c = 3(p+9)
   

2. The condition is "equal roots," which means the discriminant must be zero.

   D = b² - 4ac = 0
   

3. Substitute the coefficients carefully. Using square brackets can help avoid confusion.

   [-6(p+1)]² - 4[p+1][3(p+9)] = 0
   

4. Simplify the expression.

   36(p+1)² - 12(p+1)(p+9) = 0
   

5. Notice that 12(p+1) is a common factor. Since p ≠ -1, we know p+1 ≠ 0, so we can divide the entire equation by 12(p+1) to simplify it.

   3(p+1) - (p+9) = 0
   

6. Now, solve the simple linear equation for p.

   3p + 3 - p - 9 = 0
   

   2p - 6 = 0
   

   2p = 6
   

   p = 3
   

Final Answer: The value of p is 3.

Example 4: A Proof-based Question (Tricky)

Given: The quadratic equation (a² + b²)x² - 2(ac + bd)x + (c² + d²) = 0.

To Find: Prove that if the roots are equal, then a/b = c/d.

Solution:

1. Identify the coefficients.

   A = (a² + b²)
   B = -2(ac + bd)
   C = (c² + d²)
   

   (Using capital letters to avoid confusion with the a,b,c,d in the problem.)

2. The condition is "equal roots," so the discriminant is zero.

   D = B² - 4AC = 0
   

3. Substitute the coefficients into the discriminant formula.

   [-2(ac + bd)]² - 4(a² + b²)(c² + d²) = 0
   

4. Expand and simplify the terms carefully.

   4(ac + bd)² - 4(a²c² + a²d² + b²c² + b²d²) = 0
   

5. Divide the entire equation by 4.

   (a²c² + 2abcd + b²d²) - (a²c² + a²d² + b²c² + b²d²) = 0
   

6. The a²c² and b²d² terms cancel out when you open the bracket.

   2abcd - a²d² - b²c² = 0
   

7. Multiply by -1 and rearrange to make it look like a standard algebraic identity.

   a²d² - 2abcd + b²c² = 0
   

8. This expression is a perfect square: (ad - bc)².

   (ad - bc)² = 0
   

9. If the square of a number is zero, the number itself must be zero.

   ad - bc = 0
   

   ad = bc
   

10. Rearrange this to get the desired result. Assuming b ≠ 0 and d ≠ 0.

    a/b = c/d
    

Final Answer: We have shown that if the roots are equal, then (ad - bc)² = 0, which leads to ad = bc, and therefore a/b = c/d. Hence proved.

Tips & Tricks

Here are a few shortcuts to quickly determine the nature of roots.

Common Mistakes

Many students make small errors that lead to the wrong conclusion. Here’s what to watch out for.

Brain-Teaser Questions

Ready to test your understanding at a higher level?

1. If the equation x² + 2(k+2)x + 9k = 0 has equal roots, find the values of k.

   💡 Answer: Here a=1, b=2(k+2), c=9k. For equal roots, D=0. [2(k+2)]² - 4(1)(9k) = 0 4(k+2)² - 36k = 0 4(k² + 4k + 4) - 36k = 0 4k² + 16k + 16 - 36k = 0 4k² - 20k + 16 = 0 k² - 5k + 4 = 0 (k-4)(k-1) = 0. So, k=4 or k=1.

2. If p, q, r are real numbers and p ≠ q, show that the roots of the equation (p-q)x² + 5(p+q)x - 2(p-q) = 0 are real and unequal.

   💡 Answer: Here a = (p-q), b = 5(p+q), c = -2(p-q). Let's find the discriminant D = b² - 4ac. D = [5(p+q)]² - 4(p-q)[-2(p-q)] D = 25(p+q)² + 8(p-q)² The term (p+q)² is always ≥ 0. The term (p-q)² is also ≥ 0. Since p ≠ q, (p-q) is not zero, so (p-q)² is strictly positive (> 0). Therefore, D = (a positive number or zero) + (a strictly positive number). The result is always D > 0. Since the discriminant is always positive, the roots are always real and unequal.

3. Find the value of m such that the quadratic equation x² - 2x(1+3m) + 7(3+2m) = 0 has equal roots.

   💡 Answer: Identify coefficients: a = 1, b = -2(1+3m), c = 7(3+2m). For equal roots, D = b² - 4ac = 0. [-2(1+3m)]² - 4(1)[7(3+2m)] = 0 4(1+3m)² - 28(3+2m) = 0 Divide by 4: (1+3m)² - 7(3+2m) = 0 (1 + 6m + 9m²) - (21 + 14m) = 0 9m² - 8m - 20 = 0 Using factorization (or quadratic formula for m): 9m² - 18m + 10m - 20 = 0 9m(m-2) + 10(m-2) = 0 (9m+10)(m-2) = 0 So, m = 2 or m = -10/9.

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for your last-minute revision!