Chapter 7: Quadratic Equation
7.0 Review
A quadratic equation is a second-degree equation in one variable. It is written in the standard form \(ax^2+bx+c=0\), where \(a \neq 0\). A quadratic equation always has two values (roots) of the variable that satisfy it.
Example: A school's rectangular office room has area \(80\text{ m}^2\). If the breadth is \(x\), the length is \(x+2\).
Length \((l) = x+2 = 8+2 = 10\text{ m}\), Breadth \((b) = x = 8\text{ m}\)
A quadratic equation is a second-degree equation of one variable, of the form \(ax^2+bx+c=0\), where \(a \neq 0\). It has two values of the variable satisfying it.
7.1 Solving Quadratic Equation
(a) Factorization Method
Activity: The area of a rectangular playground is \(300\text{ m}^2\). Its length is \(1\text{ m}\) more than double its breadth.
Breadth \(=12\text{ m}\), length \(=2(12)+1=25\text{ m}\)
Worked Example 1 (Textbook)
Solve the following equations by factorization and verify each solution.
Therefore, the roots of \(x^2 + 4x=0\) are \(x = -4\) and \(x = 0\).
Therefore, the roots of \(x^2 + 6x + 8=0\) are \(x = -4\) and \(x = -2\).
Therefore, the roots of \(x^2 - 5x + 6=0\) are \(x = 2\) and \(x = 3\).
Therefore, the roots of \(x^2 - x - 6=0\) are \(x = -2\) and \(x = 3\).
Therefore, the roots of \(2x^2 + 7x + 6=0\) are \(x = -2\) and \(x = - \frac{3}{2}\).
(b) Solving Quadratic Equation by Completing the Square
Activity: Solve \(x^2-9=0\) and \(x^2-5x+6=0\) by completing the square.
Therefore, the roots of \(x^2 - 9=0\) are \(x = 3\) and \(x = -3\).
Therefore, the roots of \(x^2 - 5x + 6=0\) are \(x = 3\) and \(x = 2\).
The roots of a quadratic equation of the form \(x^2=a^2\) are \(x = \pm a\).
Worked Example 2 (Textbook)
Solve by completing the square.
Therefore, the roots of \(x^2 - 10x + 16=0\) are \(x = 8\) and \(x = 2\).
Therefore, the roots of \(x^2 - 7x + 12=0\) are \(x = 4\) and \(x = 3\).
Therefore, the roots of \(2x^2 - 7x + 6=0\) are \(x = 2\) and \(x = \frac{3}{2}\).
(c) Solving Quadratic Equation by Using Formula
We derive a general formula to solve any quadratic equation \(ax^2+bx+c=0\), \(a \neq 0\), by completing the square.
The expression \(b^2-4ac\) is called the discriminant of the quadratic equation. It determines the nature of the roots.
Worked Example 3 (Textbook)
Solve the given quadratic equations by using the formula.
Therefore, the roots of \(x^2 - 5x + 6=0\) are \(x = 3\) and \(x = 2\).
Therefore, the roots of \(49x^2 - 14x - 3=0\) are \(x = \frac{3}{7}\) and \(x = - \frac{1}{7}\).
Comparing the Three Methods
| Method | When to use | Key idea |
|---|---|---|
| Factorization | When \(ax^2+bx+c\) splits easily into two linear factors | Split the middle term so the product of the two parts equals \(ac\) |
| गुणनखण्ड | जब \(ax^2+bx+c\) सजिलै दुई रेखीय गुणनखण्डमा बाँडिन्छ | बीचको पदलाई यसरी विभाजन गर्ने कि दुई भागको गुणनफल \(ac\) बराबर होस् |
| Completing the square | Works for any quadratic; useful to derive the formula | Rewrite as \((x+k)^2 = \text{constant}\) |
| वर्ग पूरा गर्ने | जुनसुकै द्विघातका लागि लागू हुन्छ; सूत्र निकाल्न उपयोगी | \((x+k)^2 = \text{स्थिरांक}\) रूपमा लेख्ने |
| Formula method | Works for every quadratic equation, including irrational/complicated roots | Apply \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) directly |
| सूत्र विधि | हरेक द्विघात समीकरणका लागि लागू हुन्छ, अपरिमेय मूलका लागि पनि | सिधै \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) प्रयोग गर्ने |
Exercise 7.1 — Practice (Textbook)
The following are the textbook Exercise 7.1 questions. Full step-by-step solutions for these are provided in the companion Solved Questions file; final answers are given here for quick reference.
- 1Identify which given equations are quadratic equations, with reason: (a) \((x-2)^2+1=2x-3\) (b) \(x(x+1)+8=(x+2)(x-2)\) (c) \(x(2x+3)=x^2+1\) (d) \((x+2)^3=x^3-4\) (e) \(x^2+3x+1=(x-2)^2\) (f) \((x+2)^3=2x(x^2-1)\)
- 2Solve by factorization: (a) \(x^2-3x-10=0\) (b) \(2x^2+x-6=0\) (c) \(2x^2-x+\frac{1}{8}=0\) (d) \(100x^2-20x+1=0\) (e) \(x^2-45x+324=0\) (f) \(x^2-27x-182=0\)
- 3Solve by completing the square: (a) \(x^2-6x+9=0\) (b) \(9x^2-15x+6=0\) (c) \(2x^2-5x+3=0\) (d) \(5x^2-6x-2=0\) (e) \(x^2+\frac{15}{16}=2x\) (f) \(x^2+\frac{2}{3}x=\frac{35}{9}\)
- 4Solve by using formula: (a) \(x^2-9x+20=0\) (b) \(x^2+2x-143=0\) (c) \(3x^2-5x+2=0\) (d) \(2x^2-2\sqrt2x+1=0\) (e) \(x+\frac1x=3\) (f) \(\frac1x+\frac1{x-2}=3\) (g) \(\frac1{x+4}-\frac1{x-7}=\frac{11}{30}\)
- 5Ramnaresh Mahato scored a total of 30 marks in VR English and Mathematics. If he scored 2 more marks in Mathematics and 3 fewer in English, the product of the marks would be 210. Find his scores in both subjects.
- 6A rectangular playground's longer side is 30 m more than its shorter side, and its diagonal is 60 m more than its shorter side. (a) Find the length and breadth. (b) How many 12 m × 3 m turfs are needed? (c) Find the fencing cost at Rs. 15/m for 4 rounds.
| Q | Answer |
|---|---|
| 1 | (a) Yes (b) No (c) Yes (d) Yes (e) No (f) No |
| 2 | (a) \(5,-2\) (b) \(-2,\frac32\) (c) \(\frac14,\frac14\) (d) \(\frac1{10},\frac1{10}\) (e) \(9,36\) (f) \(13,14\) |
| 3 | (a) \(3,3\) (b) \(1,\frac23\) (c) \(1,\frac32\) (d) \(\frac{3+\sqrt{19}}5,\frac{3-\sqrt{19}}5\) (e) \(\frac34,\frac54\) (f) \(\frac53,-\frac73\) |
| 4 | (a) \(4,5\) (b) \(11,-13\) (c) \(1,\frac23\) (d) \(\frac1{\sqrt2},\frac1{\sqrt2}\) (e) \(\frac{3+\sqrt5}2,\frac{3-\sqrt5}2\) (f) \(\frac{4+\sqrt{10}}3,\frac{4-\sqrt{10}}3\) (g) \(1,2\) |
| 5 | 12, 18 (or 13, 17) |
| 6 | (a) 120 m, 90 m (b) 300 turfs (c) Rs. 25,200 |
7.2 Word Problems Related to Quadratic Equation
To solve a word problem using a quadratic equation: (1) assign a variable to the unknown, (2) translate the condition(s) into an equation, (3) solve the quadratic equation, and (4) reject any root that does not fit the real-world condition (e.g. negative age, negative length).
Activity 4: Ages
Sumitra's age is 12 years and her sister's is 18 years now. In how many years will the product of their ages be 280?
2 years later, the product of their ages will be 280.
Worked Example 4 (Textbook)
The two positive numbers are 7 and 11.
Worked Example 5 (Textbook)
The required positive integer is 7.
Worked Example 6 (Textbook)
The required two positive numbers are 4 and 6.
Worked Example 7 (Textbook)
The required numbers are 5 and \(\frac15\).
Worked Example 8 (Textbook)
The elder brother's age is 18 and the younger brother's age is 16.
Worked Example 9 (Textbook)
The required number is 36.
Worked Example 10 (Textbook)
8 years ago, the product of the father's and son's ages was 272.
Worked Example 11 (Textbook)
The hypotenuse of a right-angled triangle is 13 m. If the difference of its other two sides is 7 m, find the length of the remaining sides.
The remaining sides are 5 m and 12 m.
Worked Example 12 (Textbook)
The area of a rectangular land is \(50\text{ m}^2\) and its perimeter is 90 m. If the land is to be made square, by what percentage should the length be reduced?
The length and breadth are 25 m and 20 m; the length must be reduced by 20%.
Worked Example 13 (Textbook)
(a) 15 students attended. (b) Each paid Rs. 2800.
Exercise 7.2 — Practice (Textbook)
The following are the textbook Exercise 7.2 word-problem questions (19 questions, several with multiple parts). Full step-by-step solutions are provided in the companion Solved Questions file; final answers are given here for quick reference.
- 1If 11 is added to the square of a natural number, the sum is 36. Find the number.
- 2If 11 is subtracted from the square of a number, the remainder is 25. Find the number.
- 3If 7 is subtracted from double the square of a positive number, the remainder is 91. Find the number.
- 4If 2 is subtracted from the square of a natural number, the remainder is 7. Find the number.
- 5If 11 is subtracted from the square of a number and the remainder is 89, find that number.
- 6If 17 is subtracted from the square of a number, the remainder is 55. Find the number.
- 7If 3 is subtracted from double the square of a positive number, the remainder is 285. Find the number.
- 8If the sum of a number and its square is 72, find the number.
- 9If the product of two consecutive even numbers is 80, find the numbers.
- 10If the product of two consecutive odd numbers is 225, find the numbers.
- 11If the sum of a number and its reciprocal is \(\frac{10}{3}\), find the number.
- 12If the sum of two natural numbers is 21 and the sum of their squares is 261, find the numbers.
- 13If the age difference between two brothers is 4 years and the product of their ages is 221, find their ages.
- 14The sum of the present ages of two brothers is 22 and the product of their ages is 120. Find their present ages.
- 15The age difference between two sisters is 3 years and the product of their ages is 180. Find their present ages.
- 16(a) A father (40) and son (13): find how many years ago the product of their ages was 198. (b) A mother (34) and daughter (4): find how many years later the product will be 400. (c) A father (35) and son (1): find how many years later the product will be 240. (d) A husband (35) and wife (27): find how many years ago the product was 425.
- 17(a) Hypotenuse 25 m, difference of other sides 17 m: find the remaining sides. (b) Hypotenuse is double and 6 m more than the shortest side, other side is 2 m less than hypotenuse: find all sides. (c) Rectangular land area \(150\text{ m}^2\), perimeter 50 m: find length & breadth. (d) Rectangular land area \(54\text{ m}^2\), perimeter 30 m: find length & breadth. (e) Rectangle length 24 m, diagonal 16 m more than breadth: find the area.
- 18A two-digit number equals four times the sum of its digits and three times the product of its digits. Find the number.
- 19An institute planned to distribute 180 pencils equally among grade-one students. On distribution day 5 students were absent, so each student got 3 more pencils. (a) How many students were enrolled? (b) How many pencils did each student receive in total?
| Q | Answer |
|---|---|
| 1 | 5 |
| 2 | \(\pm6\) |
| 3 | 7 |
| 4 | 3 |
| 5 | \(\pm10\) |
| 6 | \(\pm6\) |
| 7 | 12 |
| 8 | 8 |
| 9 | 8 and 10 (or -10 and -8) |
| 10 | 3 and 5 (or -5 and -3) |
| 11 | 3 and \(\frac13\) |
| 12 | 6 and 15 |
| 13 | 17 years and 13 years |
| 14 | 12 years and 10 years |
| 15 | 15 years and 12 years |
| 16 | (a) 7 years (b) 6 years (c) 5 years (d) 10 years |
| 17 | (a) 24 m and 7 m (b) 10 m, 24 m, 26 m (c) 15 m and 10 m (d) 9 m and 6 m (e) 240 m\(^2\) |
| 18 | 24 |
| 19 | (a) 20 students (b) 12 pencils |
Important Theorems & Formulas
| Concept | Formula |
|---|---|
| General form of a quadratic equation | \(ax^2+bx+c=0,\ a\neq0\) |
| द्विघात समीकरणको साधारण रूप | \(ax^2+bx+c=0,\ a\neq0\) |
| Quadratic formula | \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) |
| द्विघात सूत्र | \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) |
| Discriminant | \(D=b^2-4ac\) |
| विवेचक | \(D=b^2-4ac\) |
| Roots of \(x^2=a^2\) | \(x=\pm a\) |
| \(x^2=a^2\) का मूलहरू | \(x=\pm a\) |
| Pythagoras theorem (used in right-triangle word problems) | \(h^2=p^2+b^2\) |
| पाइथागोरस प्रमेय (समकोण त्रिभुज समस्यामा प्रयोग) | \(h^2=p^2+b^2\) |
| Perimeter and area of rectangle | \(P=2(l+b)\), \(A=l\times b\) |
| आयतको परिमिति र क्षेत्रफल | \(P=2(l+b)\), \(A=l\times b\) |
Common Mistakes
- Forgetting to check \(a\neq0\) before calling an equation quadratic.
- Sign errors while moving terms across the equals sign or while splitting the middle term.
- Taking only one root and forgetting the \(\pm\) sign in completing the square / formula method.
- Accepting an impossible root (negative age, negative length, non-integer digit) instead of rejecting it based on the real-world context.
- Errors in the discriminant calculation \(b^2-4ac\), especially sign of \(4ac\).
- Forgetting units (m, m², years, Rs.) in the final answer of a word problem.
Exam Tips
- Always write the equation in standard form \(ax^2+bx+c=0\) before choosing a method.
- In word problems, clearly define the variable first, then form the equation.
- Show every step — factorization split, or the completing-the-square step, or the formula substitution.
- Always test both roots against the real-world condition and reject the impossible one with a reason.
- Double-check arithmetic under the square root before simplifying.
Quick Revision
Key Definitions
- A quadratic equation is a second-degree equation in one variable, \(ax^2+bx+c=0,\ a\neq0\), with two roots.
Key Formulas
- Quadratic formula: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
- Roots of \(x^2=a^2\): \(x=\pm a\)
Key Theorems
- The quadratic formula is derived by completing the square on \(ax^2+bx+c=0\).
Important Methods
- Factorization method
- Completing the square method
- Formula method
Important Question Patterns
- Direct factorization/completing-square/formula solving
- Number-based problems (sum/product/reciprocal of numbers)
- Age-based problems (present, past, future product/sum of ages)
- Two-digit number (digits) problems
- Right-triangle (Pythagoras) problems
- Rectangle area–perimeter problems
- Equal-sharing / distribution word problems