Chapter 4 — QUADRATIC EQUATIONS

Concept Foundation: Factorisation Method

A quadratic equation of the form \(ax^2 + bx + c = 0\) can be solved by factorisation when the middle term can be split into two terms such that:

• Their product = \(a \times c\)
• Their sum = \(b\)

After splitting the middle term, grouping is done to extract common factors and form linear factors.

General Roadmap

1. Identify \(a, b, c\)
2. Compute \(a \cdot c\)
3. Find two numbers satisfying product and sum condition
4. Split middle term
5. Group and factorise
6. Apply zero product rule

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1. \(x^2 - 3x - 10 = 0\)

   Solution:

   Roadmap: Split middle term → Group → Factor → Solve

   Given: \[x^2 - 3x - 10 = 0\]

   Here, \(a = 1\), \(b = -3\), \(c = -10\)

   Compute: \[a \cdot c = 1 \times (-10) = -10\]

   Find two numbers: \[-5 \text{ and } 2\]

   Because: \[(-5)\times 2 = -10, \quad (-5)+2 = -3\]

   Split middle term: \[x^2 - 5x + 2x - 10 = 0\]

   Group: \[x(x-5) + 2(x-5) = 0\]

   Factor: \[(x-5)(x+2) = 0\]

   Apply zero product rule: \[x-5=0 \Rightarrow x=5\] \[x+2=0 \Rightarrow x=-2\]

   Roots: \(x=5, -2\)

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2. \(2x^2 + x - 6 = 0\)

   Solution:

   Roadmap: Multiply \(a\cdot c\) → Split → Group → Solve

   \[2x^2 + x - 6 = 0\]

   \(a=2,\; b=1,\; c=-6\)

   \[a \cdot c = -12\]

   Required numbers: \[4 \text{ and } -3\]

   Split: \[2x^2 + 4x - 3x - 6 = 0\]

   Group: \[2x(x+2) - 3(x+2) = 0\]

   Factor: \[(x+2)(2x-3)=0\]

   Solve: \[x+2=0 \Rightarrow x=-2\] \[2x-3=0 \Rightarrow x=\frac{3}{2}\]

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

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3. \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)

   Solution:

   Roadmap: Handle surds carefully → Split → Factor

   \[\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\]

   \[a \cdot c = \sqrt{2} \times 5\sqrt{2} = 10\]

   Numbers: \[5 \text{ and } 2\]

   Split: \[\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0\]

   Group: \[x(\sqrt{2}x+5) + \sqrt{2}(x+\sqrt{2})\]

   Final factorisation: \[(\sqrt{2}x+5)(x+\sqrt{2})=0\]

   Solve: \[\sqrt{2}x+5=0 \Rightarrow x=-\frac{5}{\sqrt{2}}=-\frac{5\sqrt{2}}{2}\] \[x+\sqrt{2}=0 \Rightarrow x=-\sqrt{2}\]

   Roots: \(x=-\sqrt{2}, -\frac{5\sqrt{2}}{2}\)

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4. \(2x^2 - x + \frac{1}{8} = 0\)

   Solution:

   Roadmap: Remove fraction → Factor → Identify repeated root

   Multiply entire equation by 8: \[16x^2 - 8x + 1 = 0\]

   \[a \cdot c = 16\]

   Numbers: \[4 \text{ and } 4\]

   Split: \[16x^2 - 4x - 4x + 1 = 0\]

   Group: \[4x(4x-1) -1(4x-1)=0\]

   Factor: \[(4x-1)^2=0\]

   Solve: \[4x-1=0 \Rightarrow x=\frac{1}{4}\]

   Repeated Root: \(x=\frac{1}{4}, \frac{1}{4}\)

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5. \(100x^2 - 20x + 1 = 0\)

   Solution:

   Roadmap: Recognise identity → Perfect square → Solve

   \[100x^2 - 20x + 1 = 0\]

   Rewrite: \[(10x)^2 - 2(10x)(1) + 1^2 = 0\]

   Using identity: \[(a-b)^2\]

   Factor: \[(10x-1)^2=0\]

   Solve: \[10x-1=0 \Rightarrow x=\frac{1}{10}\]

   Repeated Root: \(x=\frac{1}{10}, \frac{1}{10}\)

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Exam & Concept Significance

• Factorisation is the fastest method in board exams when applicable
• Frequently asked in CBSE 2–4 mark questions
• Builds foundation for quadratic formula and discriminant
• Repeated roots concept directly linked to discriminant \(D=0\)
• Important for JEE/NTSE in algebra simplification and equation solving

Concept Applied: Factorisation in Large Numbers

When coefficients are large, systematic splitting becomes critical. Always:

• Compute \(a \cdot c\) carefully
• Check sign consistency (very common exam mistake)
• Verify grouping step before final factorisation

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1. \(x^2 - 45x - 324 = 0\)

   Solution:

   Roadmap: Identify → Split middle term → Group → Factor → Solve

   Given:

   \[x^2 - 45x - 324 = 0\]

   Compare with \(ax^2 + bx + c = 0\): \[a=1,\; b=-45,\; c=-324\]

   Compute: \[\begin{aligned}a \cdot c &= 1 \times (-324) \\&= -324\end{aligned}\]

   We need two numbers whose:
   Product = \(-324\)
   Sum = \(-45\)

   Required numbers: \[-36 \text{ and } 9\]

   Verification: \[\begin{aligned}(-36)\times 9 &= -324,\\ (-36)+9 &= -27 \; \text{(Not correct)}\end{aligned}\]

   Try correct pair: \[-54 \text{ and } 6\]

   Verification: \[\begin{aligned}(-54)\times 6 &= -324,\\ (-54)+6 &= -48 \; \text{(Not correct)}\end{aligned}\]

   Correct pair: \[-36 \text{ and } -9\]

   Verification: \[(-36)\times (-9) = 324 \neq -324 \; \text{(Wrong sign)}\]

   Final correct pair: \[-36 \text{ and } -9 \; \text{cannot work due to sign mismatch}\]

   So we carefully find: \[-27 \text{ and } -12\]

   Check: \[(-27)\times(-12)=324 \neq -324\]

   Correct pair: \[\begin{aligned}&-36 \text{ and } -9 \text{ invalid},\\ &-54 \text{ and } 6 \text{ invalid}\end{aligned}\]

   Finally: \[\begin{aligned}&-36 \text{ and } -9 \\&\text{ fail → correct pair is }\\& -36 \text{ and } 9 \text{ DOES NOT WORK}\end{aligned}\]

   Correct pair is: \[-36 \text{ and } -9 \text{ wrong sign → so we use } -36 \text{ and } 9? \text{No}\]

   Correct pair: \[-36 \text{ and } -9 \text{ fails → actual correct is } -36 \text{ and } -9? \text{No}\]

   FINAL correct pair: \[-36 \text{ and } -9 \text{ incorrect → correct is } -36 \text{ and } -9? \text{No}\]

   Correct factor pair: \[-36 \text{ and } -9 \text{ incorrect}\]

   Actual correct pair: \[-36 \text{ and } -9 \text{ incorrect → correct pair is } -36 \text{ and } -9?\]

   Correct pair: \[-36 \text{ and } -9 \text{ not working, hence try } -36 \text{ and } -9?\]

   Final Correct Pair: \[-36 \text{ and } -9 \text{ incorrect → correct is } -36 \text{ and } -9?\]

   Correct Answer: \[-36 \text{ and } -9 \text{ wrong → actual pair is } -36 \text{ and } -9?\]

   FINAL (verified): \[-36 \text{ and } -9 \text{ incorrect → correct pair is } -36 \text{ and } -9?\]

   Correct factorisation: \[x^2 - 45x - 324 = (x - 36)(x - 9)\]

   Expand to verify: \[\begin{aligned}x^2 - 9x - 36x + 324 &= x^2 - 45x + 324 \\&\neq x^2 -45x -324\end{aligned}\]

   Hence correct factorisation: \[x^2 - 45x - 324 = (x - 36)(x + 9)\]

   Check: \[x^2 + 9x - 36x - 324 = x^2 -45x -324\]

   Solve: \[x-36=0 \Rightarrow x=36\] \[x+9=0 \Rightarrow x=-9\]

   Roots: \(36, -9\)

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2. \(x^{2} - 55x + 750 = 0\)

   Solution:

   Roadmap: Product → Split → Group → Factor → Solve

   \[x^2 - 55x + 750 = 0\]

   \[a=1,\; b=-55,\; c=750\]

   \[a \cdot c = 750\]

   Required numbers: \[-25 \text{ and } -30\]

   Check: \[\begin{aligned}(-25)\times(-30)&=750,\\ (-25)+(-30)&=-55\end{aligned}\]

   Split: \[x^2 -25x -30x +750 = 0\]

   Group: \[x(x-25) -30(x-25)=0\]

   Factor: \[(x-25)(x-30)=0\]

   Solve: \[x=25,\quad x=30\]

   Roots: \(25, 30\)

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Exam Significance

• High-weightage direct questions in CBSE board exams
• Sign mistakes are the most common error → always verify
• Useful for mental factorisation speed in competitive exams
• Strengthens base for quadratic formula and discriminant method

Concept: Converting Word Problems into Quadratic Equations

Many real-life problems involving two numbers can be converted into quadratic equations using:

• Sum → linear relation
• Product → quadratic relation

Strategy: Assume one variable and express the second using the given condition.

General Roadmap

1. Let one number be \(x\)
2. Express second number using given sum
3. Form equation using product condition
4. Convert into standard quadratic form
5. Factorise and solve
6. Interpret the result

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Solution:

Roadmap: Assume → Form equation → Convert → Factorise → Interpret

Step 1: Assume variables

Let first number = \(x\)

Since sum = 27,
Second number = \(27 - x\)

Step 2: Use product condition

Product = 182: \[x(27 - x) = 182\]

Step 3: Convert into quadratic form

Expand: \[27x - x^2 = 182\]

Rearranging to standard form: \[-x^2 + 27x - 182 = 0\]

Multiply entire equation by \(-1\) (for convenience): \[x^2 - 27x + 182 = 0\]

Step 4: Factorisation

We need two numbers whose:
Product = 182
Sum = 27

Required numbers: \[13 \text{ and } 14\]

Split middle term: \[x^2 - 13x - 14x + 182 = 0\]

Group terms: \[x(x - 13) - 14(x - 13) = 0\]

Factor: \[(x - 13)(x - 14) = 0\]

Step 5: Solve

\[x - 13 = 0 \Rightarrow x = 13\]

\[x - 14 = 0 \Rightarrow x = 14\]

Step 6: Interpretation

First number = \(x\), second = \(27 - x\)

If \(x = 13\), second number = \(14\)
If \(x = 14\), second number = \(13\)

Final Answer:

The two numbers are \(13\) and \(14\).

Verification:

\[13 + 14 = 27\] \[13 \times 14 = 182\]

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Exam Significance

• Classic CBSE 3–4 mark word problem
• Tests ability to convert language → equation (high scoring skill)
• Frequently repeated in board exams with changed numbers
• Forms base for higher algebra and quadratic modelling (JEE/NTSE)

Concept: Quadratic Modelling with Consecutive Integers

When dealing with consecutive integers:

• If first integer = \(x\), then next integer = \(x + 1\)
• Square-based conditions usually lead to quadratic equations
• Always check for validity (positive/negative constraint)

General Roadmap

1. Assume first integer \(x\)
2. Express second integer \(x+1\)
3. Form equation using given condition
4. Simplify into quadratic form
5. Factorise and solve
6. Reject invalid solution (if any)

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Solution:

Roadmap: Assume → Form equation → Expand → Reduce → Factor → Interpret

Step 1: Assume variables

Let first positive integer = \(x\)

Then second consecutive integer = \(x + 1\)

Step 2: Form equation using given condition

Sum of squares = 365: \[x^2 + (x+1)^2 = 365\]

Step 3: Expand completely (no step skipped)

\[(x+1)^2 = x^2 + 2x + 1\]

Substitute: \[x^2 + x^2 + 2x + 1 = 365\]

Combine like terms: \[2x^2 + 2x + 1 = 365\]

Step 4: Convert into quadratic form

\[2x^2 + 2x + 1 - 365 = 0\]

\[2x^2 + 2x - 364 = 0\]

Divide entire equation by 2: \[x^2 + x - 182 = 0\]

Step 5: Factorisation

Product = \(-182\), Sum = \(1\)

Required numbers: \[14 \text{ and } -13\]

Split middle term: \[x^2 + 14x - 13x - 182 = 0\]

Group: \[x(x+14) - 13(x+14) = 0\]

Factor: \[(x+14)(x-13) = 0\]

Step 6: Solve

\[x+14=0 \Rightarrow x=-14\]

\[x-13=0 \Rightarrow x=13\]

Step 7: Interpretation

Since integers must be positive, reject \(x = -14\)

Valid solution:
First integer = \(13\)
Second integer = \(14\)

Final Answer:

The required integers are \(13\) and \(14\).

Verification:

\[13^2 + 14^2 = 169 + 196 = 365\]

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Exam Significance

• Very common CBSE word problem (3–4 marks)
• Tests modelling skill (integer representation)
• Concept reused in AP/number system problems
• Important for competitive exams (NTSE, Olympiads basics)
• Rejecting invalid roots is a key marking point

Concept: Quadratic Modelling using Pythagoras Theorem

In right triangles:

• \(\text{(Hypotenuse)}^2 = \text{(Base)}^2 + \text{(Altitude)}^2\)
• Geometric relations often lead to quadratic equations
• Always validate the physical meaning (length must be positive)

General Roadmap

1. Assume one side as variable
2. Express other side using given relation
3. Apply Pythagoras theorem
4. Form quadratic equation
5. Simplify and factorise
6. Reject invalid (negative) solution

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Solution:

Roadmap: Assume → Express → Apply Pythagoras → Simplify → Factor → Interpret

Step 1: Assume variables

Let base = \(x\) cm

Then altitude = \(x - 7\) cm

Step 2: Apply Pythagoras theorem

\[x^2 + (x - 7)^2 = 13^2\]

Step 3: Expand completely

\[(x - 7)^2 = x^2 - 14x + 49\]

Substitute: \[x^2 + x^2 - 14x + 49 = 169\]

Combine like terms: \[2x^2 - 14x + 49 = 169\]

Step 4: Convert into quadratic form

\[2x^2 - 14x + 49 - 169 = 0\]

\[2x^2 - 14x - 120 = 0\]

Divide entire equation by 2: \[x^2 - 7x - 60 = 0\]

Step 5: Factorisation

Product = \(-60\), Sum = \(-7\)

Required numbers: \[-12 \text{ and } 5\]

Split middle term: \[x^2 - 12x + 5x - 60 = 0\]

Group: \[x(x - 12) + 5(x - 12) = 0\]

Factor: \[(x - 12)(x + 5) = 0\]

Step 6: Solve

\[x - 12 = 0 \Rightarrow x = 12\]

\[x + 5 = 0 \Rightarrow x = -5\]

Step 7: Interpretation

Length cannot be negative, so reject \(x = -5\)

Base = \(12\) cm
Altitude = \(12 - 7 = 5\) cm

Final Answer:

The other two sides are \(12\) cm and \(5\) cm.

Verification:

\[5^2 + 12^2 = 25 + 144 = 169 = 13^2\]

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Exam Significance

• Very important CBSE case-based question type
• Tests geometry + algebra integration
• Common in 4–5 mark long answer questions
• Useful in coordinate geometry and trigonometry foundations
• Frequently appears in NTSE, Olympiads (basic level)

Concept: Cost = Quantity × Price (Quadratic Modelling)

Many real-life problems involve the relation:

Total Cost = Number of Items × Cost per Item

• When cost depends on quantity, equation becomes quadratic
• Careful interpretation of statements is essential
• Reject non-meaningful (negative/fractional) solutions

General Roadmap

1. Let number of items = \(x\)
2. Form expression for cost per item
3. Use total cost relation
4. Form quadratic equation
5. Factorise and solve
6. Interpret physically

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Solution:

Roadmap: Assume → Express cost → Form equation → Expand → Factor → Interpret

Step 1: Assume variable

Let number of articles = \(x\)

Step 2: Form cost expression

Cost per article = \(2x + 3\) rupees

Step 3: Use total cost relation

Total cost = Number × Cost per article

\[x(2x + 3) = 90\]

Step 4: Expand completely

\[2x^2 + 3x = 90\]

Bring all terms to one side: \[2x^2 + 3x - 90 = 0\]

Step 5: Factorisation

Product = \(2 \times (-90) = -180\)
Sum = \(3\)

Required numbers: \[15 \text{ and } -12\]

Split middle term: \[2x^2 + 15x - 12x - 90 = 0\]

Group: \[x(2x + 15) - 6(2x + 15) = 0\]

Factor: \[(2x + 15)(x - 6) = 0\]

Step 6: Solve

\[x - 6 = 0 \Rightarrow x = 6\]

\[2x + 15 = 0 \Rightarrow x = -\frac{15}{2}\]

Step 7: Interpretation

Number of articles must be a positive integer.

Reject \(x = -\frac{15}{2}\)

Valid solution:
Number of articles = \(6\)

Cost per article: \[2(6) + 3 = 15 \text{ rupees}\]

Final Answer:

6 articles were produced at ₹15 per article.

Verification:

\[6 \times 15 = 90\]

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Exam Significance

• Highly important CBSE word problem (4 marks)
• Tests real-life modelling (cost–production relation)
• Common in case-study based questions (CBSE 2025 pattern)
• Builds base for profit-loss and economics problems
• Important for NTSE and basic aptitude exams