NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.1 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.1 are provided here to assist students in their Class 10 exam preparations. These solutions are created by subject experts to help students understand the concept of quadratic equations, how to identify and solve quadratic equations using factorization and the quadratic formula, and how to apply these techniques to solve problems effectively.

Exercise 4.1 in Quadratic Equations introduces students to the basic concept of quadratic equations in the form of

NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.1 Quadratic Equations

Quadratic Equations

Medium

Q.

$\begin{array}{l}\text{Check whether the following are quadratic equations :}\\ \left(\text{i}\right){(\mathrm{x}+1)}^2=2(\mathrm{x}-3)\left(\text{ii}\right){\mathrm{x}}^2-2\mathrm{x}=-2(3-\mathrm{x})\\ \left(\mathrm{iii}\right)(\mathrm{x}-2)(\mathrm{x}+1)=(\mathrm{x}-1)(\mathrm{x}+3)\left(\text{iv}\right)(\mathrm{x}-3)(2\mathrm{x}+1)=\mathrm{x}(\mathrm{x}+5)\\ \left(\mathrm{v}\right)(2\mathrm{x}-1)(\mathrm{x}-3)=(\mathrm{x}+5)(\mathrm{x}-1)\left(\text{vi}\right){\text{ x}}^2+3\mathrm{x}+1={(\mathrm{x}-2)}^2\\ \left(\mathrm{vii}\right){(\mathrm{x}+2)}^3=2\mathrm{x}({\mathrm{x}}^2-1)\left(\text{viii}\right){\text{ x}}^3-4{\mathrm{x}}^2-\mathrm{x}+1={(\mathrm{x}-2)}^3\end{array}$

View Solution

Quadratic Equations

Medium

Q.

Represent the following situations in the form of quadratic equations:

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

2. The product of two consecutive positive integers is 306. We need to find the  integers.

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

4. 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. We need to find the speed of the train.

View Solution

$ax^2 + bx + c = 0$

ax2+bx+c=0. The exercise helps students practice problems related to identifying quadratic equations, finding roots using factorization, and exploring the nature of roots. The solutions are step-by-step, following the CBSE guidelines, and help in developing problem-solving skills in algebra.

These solutions are aligned with the latest CBSE syllabus and provide detailed, easy-to-understand steps that are essential for Class 10 exams.

Q1. Check whether the following are quadratic equations

A quadratic equation is in the standard form ax² + bx + c = 0, where a ≠ 0 and the degree is 2.

1. $(x+1)^2 = 2(x−3)$(x+1)2=2(x−3)
   Solution:
   Expanding both sides:

   $(x + 1)^2 = x^2 + 2x + 1$(x+1)2=x2+2x+1

   $2(x − 3) = 2x − 6$2(x−3)=2x−6
   Now, equating both sides:

   $x^2 + 2x + 1 = 2x − 6$x2+2x+1=2x−6
   Simplify:

   $x^2 + 2x + 1 − 2x + 6 = 0$x2+2x+1−2x+6=0

   $x^2 + 7 = 0$x2+7=0
   This is a quadratic equation, as the degree is 2.
   Quadratic equation:

   $x^2 + 7 = 0$x2+7=0

2. $x^2−2x = (−2)(3−x)$x2−2x=(−2)(3−x)
   Solution:
   Expanding both sides:

   $x^2 − 2x = -6 + 2x$x2−2x=−6+2x
   Bring all terms to one side:

   $x^2 − 2x − 2x + 6 = 0$x2−2x−2x+6=0

   $x^2 − 4x + 6 = 0$x2−4x+6=0
   This is a quadratic equation, as the degree is 2.
   Quadratic equation:

   $x^2 − 4x + 6 = 0$x2−4x+6=0

3. $(x−2)(x+1) = (x−1)(x+3)$(x−2)(x+1)=(x−1)(x+3)
   Solution:
   Expanding both sides:

   $(x − 2)(x + 1) = x^2 + x − 2x − 2 = x^2 − x − 2$(x−2)(x+1)=x2+x−2x−2=x2−x−2

   $(x − 1)(x + 3) = x^2 + 3x − x − 3 = x^2 + 2x − 3$(x−1)(x+3)=x2+3x−x−3=x2+2x−3
   Now, equating both sides:

   $x^2 − x − 2 = x^2 + 2x − 3$x2−x−2=x2+2x−3
   Simplify:

   $-x − 2 − 2x + 3 = 0$−x−2−2x+3=0

   $-3x + 1 = 0$−3x+1=0
   This is a linear equation, not quadratic (degree = 1).
   Not a quadratic.

4. $(x−3)(2x+1) = x(x+5)$(x−3)(2x+1)=x(x+5)
   Solution:
   Expanding both sides:

   $(x − 3)(2x + 1) = 2x^2 + x − 6x − 3 = 2x^2 − 5x − 3$(x−3)(2x+1)=2x2+x−6x−3=2x2−5x−3

   $x(x + 5) = x^2 + 5x$x(x+5)=x2+5x
   Now, equating both sides:

   $2x^2 − 5x − 3 = x^2 + 5x$2x2−5x−3=x2+5x
   Bring all terms to one side:

   $2x^2 − 5x − 3 − x^2 − 5x = 0$2x2−5x−3−x2−5x=0

   $x^2 − 10x − 3 = 0$x2−10x−3=0
   This is a quadratic equation, as the degree is 2.
   Quadratic equation:

   $x^2 − 10x − 3 = 0$x2−10x−3=0

5. $(2x−1)(x−3) = (x+5)(x−1)$(2x−1)(x−3)=(x+5)(x−1)
   Solution:
   Expanding both sides:

   $(2x − 1)(x − 3) = 2x^2 − 6x − x + 3 = 2x^2 − 7x + 3$(2x−1)(x−3)=2x2−6x−x+3=2x2−7x+3

   $(x + 5)(x − 1) = x^2 − x + 5x − 5 = x^2 + 4x − 5$(x+5)(x−1)=x2−x+5x−5=x2+4x−5
   Now, equating both sides:

   $2x^2 − 7x + 3 = x^2 + 4x − 5$2x2−7x+3=x2+4x−5
   Bring all terms to one side:

   $2x^2 − 7x + 3 − x^2 − 4x + 5 = 0$2x2−7x+3−x2−4x+5=0

   $x^2 − 11x + 8 = 0$x2−11x+8=0
   This is a quadratic equation, as the degree is 2.
   Quadratic equation:

   $x^2 − 11x + 8 = 0$x2−11x+8=0

6. $x^2 + 3x + 1 = (x−2)^2$x2+3x+1=(x−2)2
   Solution:
   Expanding the right side:

   $(x − 2)^2 = x^2 − 4x + 4$(x−2)2=x2−4x+4
   Now, equating both sides:

   $x^2 + 3x + 1 = x^2 − 4x + 4$x2+3x+1=x2−4x+4
   Simplify:

   $3x + 1 + 4x − 4 = 0$3x+1+4x−4=0

   $7x − 3 = 0$7x−3=0
   This is a linear equation, not quadratic (degree = 1).
   Not a quadratic.

7. $(2x−1)(x−3) = (x+5)(x−1)$(2x−1)(x−3)=(x+5)(x−1)
   Solution:
   Expanding both sides:

   $(2x − 1)(x − 3) = 2x^2 − 6x − x + 3 = 2x^2 − 7x + 3$(2x−1)(x−3)=2x2−6x−x+3=2x2−7x+3

   $(x + 5)(x − 1) = x^2 − x + 5x − 5 = x^2 + 4x − 5$(x+5)(x−1)=x2−x+5x−5=x2+4x−5
   Now, equating both sides:

   $2x^2 − 7x + 3 = x^2 + 4x − 5$2x2−7x+3=x2+4x−5
   Bring all terms to one side:

   $2x^2 − 7x + 3 − x^2 − 4x + 5 = 0$2x2−7x+3−x2−4x+5=0

   $x^2 − 11x + 8 = 0$x2−11x+8=0
   This is a quadratic equation, as the degree is 2.
   Quadratic equation:

   $x^2 − 11x + 8 = 0$x2−11x+8=0

Q2. Represent the following situations as quadratic equations

(i) Rectangular plot area problem

Let the breadth be x.
Then, the length =

$2x + 1$

2x+1.
Area = 528 ⇒

$x(2x + 1) = 528$

x(2x+1)=528
⇒

$2x^2 + x − 528 = 0$

2x2+x−528=0
Quadratic equation:

$2x^2 + x − 528 = 0$

2x2+x−528=0

(ii) Product of two consecutive integers is 306

Let the first integer be x, and the next one is

$x + 1$

x+1.
The product = 306 ⇒

$x(x + 1) = 306$

x(x+1)=306
⇒

$x^2 + x − 306 = 0$

x2+x−306=0
Quadratic equation:

$x^2 + x − 306 = 0$

x2+x−306=0

(iii) Rohan’s mother and his age problem

Let Rohan’s present age = x.
Rohan’s mother’s present age =

$x + 26$

x+26.
After 3 years: Rohan’s age =

$x + 3$

x+3 and mother’s age =

$x + 29$

x+29.
Product of their ages = 360 ⇒

$(x + 3)(x + 29) = 360$

(x+3)(x+29)=360
⇒

$x^2 + 32x − 273 = 0$

x2+32x−273=0
Quadratic equation:

$x^2 + 32x − 273 = 0$

x2+32x−273=0

(iv) Train speed problem

Distance = 480 km.
Normal speed = s, time =

$480/s$

480/s.
If the speed is reduced by 8 km/h, the time increases by 3 hours ⇒

$480/(s−8) = 480/s + 3$

480/(s−8)=480/s+3.
Simplifying gives:

$3s^2 − 24s − 3840 = 0$

3s2−24s−3840=0
Quadratic equation:

$3s^2 − 24s − 3840 = 0$

3s2−24s−3840=0

FAQs: Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.1

Q1. What is a quadratic equation?
Answer:
A quadratic equation is an algebraic equation of the form:

$ax^2 + bx + c = 0$

ax2+bx+c=0,
where a, b, and c are constants, and

$a \neq 0$

a=0. The highest degree of the variable

$x$

x is 2.

Q2. How do I solve quadratic equations in Exercise 4.1?
Answer:
Quadratic equations can be solved using the following methods:

• Factorization Method: Express the quadratic equation as a product of two binomials and solve for

  $x$x.

• Quadratic Formula: Use the formula

  $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$x=2a−b±b2−4ac​​ to find the roots of the equation.

Q3. What is the quadratic formula?
Answer:
The quadratic formula is used to find the roots of a quadratic equation

$ax^2 + bx + c = 0$

ax2+bx+c=0. It is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

x=2a−b±b2−4ac​​

Q4. What are the roots of a quadratic equation?
Answer:
The roots of a quadratic equation are the values of

$x$

x that satisfy the equation. They are also known as the solutions of the equation.

Q5. How do NCERT Solutions help in exam preparation?
Answer:
These solutions provide step-by-step explanations for solving quadratic equations using factorization and the quadratic formula. They help students understand how to approach and solve problems systematically, ensuring better performance in Class 10 exams.