NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.2 are provided to help students with their Class 10 exam preparations. These solutions are carefully crafted by subject experts to explain the process of solving quadratic equations in detail, using methods like factorization, and the quadratic formula.

Exercise 4.2 focuses on solving word problems and applying the concepts of quadratic equations to real-life situations. By practicing this exercise, students will learn how to form quadratic equations from given conditions and solve them to find practical solutions.

NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations

Quadratic Equations
Medium
Q.
Find the roots of the following quadratic equationsby factorisation:  (i) x2−3x−10=0               (ii) 2x2+x−6=0(iii) 2x2+7x+52=0        (iv) 2x2−x+18=0(v) 100x2−20x+1=0 \begin{array}{l}{Find the roots of the following quadratic equations}\\ {by factorisation:}\\ {  (i) }{\mathrm{x}}^{2}-3\mathrm{x}-10=0{               (ii) }2{\mathrm{x}}^{2}+\mathrm{x}-6=0\\ {(iii) }\sqrt{2}{\mathrm{x}}^{2}+7\mathrm{x}+5\sqrt{2}=0{{        (iv) 2x}}^{2}-\mathrm{x}+\frac{1}{8}=0\\ {{(v) 100x}}^{2}-20\mathrm{x}+1=0\end{array}
Quadratic Equations
Medium
Q.
Find two numbers whose sum is 27 and product is 182.
Quadratic Equations
Medium
Q.
Find two consecutive positive integers, sum of whose squares is 365.
Quadratic Equations
Medium
Q.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Quadratic Equations
Medium
Q.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Quadratic Equations
Medium
Q.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Quadratic Equations
Medium
Q.
A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) Solve the quadratic equation for x, using quadratic formula.
[CBSE - 2024]

NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations

These solutions are aligned with the latest CBSE syllabus, providing step-by-step explanations to make students proficient in solving quadratic equations and ready for their Class 10 exams.

Question 1: Find the roots of the following quadratic equations by factorisation.

(i) $x^2 - 3x - 10 = 0$

  • Step: Find two numbers that multiply to $-10$ and add to $-3$. (Numbers: $-5, 2$)

  • $x^2 - 5x + 2x - 10 = 0$

  • $x(x - 5) + 2(x - 5) = 0$

  • $(x - 5)(x + 2) = 0$

  • Roots: $x = 5$ and $x = -2$.

(ii) $2x^2 + x - 6 = 0$

  • Step: Multiply $2 \times (-6) = -12$. Find numbers that multiply to $-12$ and add to $1$. (Numbers: $4, -3$)

  • $2x^2 + 4x - 3x - 6 = 0$

  • $2x(x + 2) - 3(x + 2) = 0$

  • $(2x - 3)(x + 2) = 0$

  • Roots: $x = \frac{3}{2}$ and $x = -2$.

(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

  • Step: Multiply $\sqrt{2} \times 5\sqrt{2} = 10$. Find numbers that multiply to $10$ and add to $7$. (Numbers: $5, 2$)

  • $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$

  • $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$ (Note: $2x$ can be written as $\sqrt{2} \cdot \sqrt{2}x$)

  • $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$

  • Roots: $x = -\frac{5}{\sqrt{2}}$ and $x = -\sqrt{2}$.

Question 3: Find two numbers whose sum is 27 and product is 182.

  • Solution:

    • Let the first number be $x$.

    • The second number will be $(27 - x)$.

    • According to the question: $x(27 - x) = 182$

    • $27x - x^2 = 182 \implies x^2 - 27x + 182 = 0$

    • Factoring: $x^2 - 13x - 14x + 182 = 0$

    • $x(x - 13) - 14(x - 13) = 0 \implies (x - 13)(x - 14) = 0$

    • Answer: The two numbers are 13 and 14.

Question 4: Find two consecutive positive integers, the sum of whose squares is 365.

  • Solution:

    • Let the first integer be $x$ and the next be $(x + 1)$.

    • Equation: $x^2 + (x + 1)^2 = 365$

    • $x^2 + x^2 + 2x + 1 = 365 \implies 2x^2 + 2x - 364 = 0$

    • Dividing by 2: $x^2 + x - 182 = 0$

    • $x^2 + 14x - 13x - 182 = 0 \implies (x + 14)(x - 13) = 0$

    • Since it must be a positive integer, $x = 13$.

    • Answer: The integers are 13 and 14.

Question 5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

  • Solution:

    • Let the base be $x$ cm. Then altitude $= (x - 7)$ cm.

    • By Pythagoras Theorem: $Base^2 + Altitude^2 = Hypotenuse^2$

    • $x^2 + (x - 7)^2 = 13^2$

    • $x^2 + x^2 - 14x + 49 = 169 \implies 2x^2 - 14x - 120 = 0$

    • Dividing by 2: $x^2 - 7x - 60 = 0$

    • Factoring: $x^2 - 12x + 5x - 60 = 0 \implies (x - 12)(x + 5) = 0$

    • $x = 12$ (Side length cannot be negative).

    • Answer: Base = 12 cm and Altitude = $12 - 7 =$ 5 cm.

Question 6: A cottage industry produces a certain number of pottery articles in a day...

(Summary: Total cost was ₹90. Cost of each article was 3 more than twice the number of articles.)

  • Solution:

    • Let the number of articles be $x$.

    • Cost of each article $= (2x + 3)$.

    • Total cost: $x(2x + 3) = 90 \implies 2x^2 + 3x - 90 = 0$

    • Factoring: $2x^2 + 15x - 12x - 90 = 0$

    • $x(2x + 15) - 6(2x + 15) = 0 \implies (2x + 15)(x - 6) = 0$

    • $x = 6$ (Number of articles cannot be a fraction or negative).

    • Answer: Number of articles = 6, Cost per article = ₹15.

FAQs: Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.2

Q1. What is the focus of Exercise 4.2?
Answer:
Exercise 4.2 focuses on solving word problems involving quadratic equations. The problems require forming quadratic equations from the given scenarios and solving them using methods like factorization and the quadratic formula.

Q2. How do I form a quadratic equation from a word problem?
Answer:
To form a quadratic equation from a word problem:

  1. Identify the unknowns in the problem.

  2. Translate the given information into an algebraic expression.

  3. Set up the equation in the standard form

    ax2+bx+c=0ax^2 + bx + c = 0.

  4. Solve the equation using factorization or the quadratic formula.

Q3. What methods are used to solve quadratic equations in this exercise?
Answer:
In this exercise, the following methods are used:

  • Factorization Method: Break down the quadratic expression into factors and solve for the values of

    xx.

  • Quadratic Formula: The roots of a quadratic equation

    ax2+bx+c=0ax^2 + bx + c = 0 can be found using the formula:

     

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q4. How can quadratic equations be applied in real life?
Answer:
Quadratic equations are widely used in real-life situations such as:

  • Calculating areas, e.g., the area of a rectangle.

  • Projectile motion, such as finding the height of an object thrown in the air.

  • Profit and loss problems in business and finance.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide detailed step-by-step methods for solving quadratic equations, helping students improve their conceptual understanding and problem-solving skills. By practicing these solutions, students can gain confidence and perform better in board exams.