156-NCERT-10-4-Quadratic Equations - Ex-4.1

NCERT

10th Mathematics

Exercise 4.1

Topic: 4 Quadratic Equations

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables-Ex-3.7

EXERCISE 4.1

1. Check whether the following are quadratic equations :

(i) (x + 1)² = 2(x – 3)     (ii) x² – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)     (iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)     (vi) x² + 3x + 1 = (x – 2)²

(vii) (x + 2)³ = 2x (x² – 1)     (viii) x³ – 4x² – x + 1 = (x – 2)³

Explanation:

1) The quadratic polynomial is of the form ax² + bx + c, where a ≠ 0.

2) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

3) Equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a

quadratic equation.

4) The equation of the form ax² + bx + c = 0, where a ≠ 0 is known as the standard

form of a quadratic equation. The degree of this equation is 2.

Solution:

(i) (x + 1)² = 2(x – 3)

1) The given equation is:

(x + 1)² = 2(x – 3)

x² + 2x +1 = 2x – 6

x² + 2x +1 - 2x + 6 = 0

x² + 7 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation. 

(ii) x² – 2x = (–2) (3 – x)

1) The given equation is:

x² – 2x = (–2) (3 – x)

x² – 2x = – 6 + 2x

x² – 2x + 6 – 2x = 0

x² – 4x + 6 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation. 

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

1) The given equation is:

(x – 2)(x + 1) = (x – 1)(x + 3)

x(x + 1) – 2(x + 1) = x(x + 3) – (x + 3)

x² + x – 2x – 2 = x² + 3x – x – 3

x² – x – 2 = x² + 2x – 3
x² – x – 2 – x² – 2x + 3 = 0

– 3x + 1 = 0
3x – 1 = 0 -------------------equation 1

2) As the highest power of the variable x is 1, it is not the quadratic equation. 

(iv) (x – 3)(2x +1) = x(x + 5)

1) The given equation is:

(x – 3)(2x +1) = x(x + 5)

x(2x + 1) – 3(2x + 1) = x(x + 5)

2x² + x – 6x – 3 = x² + 5x

2x² + x – 6x – 3 – x² – 5x = 0

x² – 10x – 3 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

1) The given equation is:

(2x – 1)(x – 3) = (x + 5)(x – 1)

2x(x – 3) – (x – 3) = x(x – 1) + 5(x – 1)

2x² – 6x – x + 3 = x² – x + 5x – 5

2x² – 6x – x + 3 – x² + x – 5x + 5 = 0

2x² – x² – 6x – x + x – 5x + 3 + 5 = 0

x² – 11x + 8 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

1) The given equation is:

x² + 3x + 1 = (x – 2)²

x² + 3x + 1 = x² – 4x + 4

x² + 3x + 1 – x² + 4x – 4 = 0

x² – x² + 3x + 4x + 1 – 4 = 0

7x – 3 = 0-------------------equation 1

2) As the highest power of the variable x is 1, it is not the quadratic equation.

(vii) (x + 2)³ = 2x (x² – 1)

1) We know that (a + b)³ = a³ + 3a² b + 3a b² + b³ 

(x + 2)³ = 2x (x² – 1)

x³ + 3x² (2) + 3x (2)² + 2³ = 2x³ – 2x

x³ + 6x² + 12x + 8 = 2x³ – 2x

x³ + 6x² + 12x + 8 – 2x³ + 2x = 0
x³ – 2x³ + 6x² + 12x + 2x + 8 = 0

– x³ + 6x² + 14x + 8 = 0 -------------------equation 1

2) As the highest power of the variable x is 3, it is not the quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

1) We know that (a - b)³ = a³ – 3a² b + 3a b² – b³ 

x³ – 4x² – x + 1 = (x – 2)³

x³ – 4x² – x + 1 = x³ – 3x² (2) + 3x (2)² – 2³

x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8

x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0

2x² – 13x + 9 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

Q2. Represent the following situations in the form of quadratic equations :

(i) 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.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) 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.

(iv) 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.

Explanation:

1) Here, let x be any suitable variable.

2) According to the given equation, frame the quadratic equation.

3) Then solve this equation to get the required solutions. 

Solution:

(i) 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.

1) Here, length is depending on breadth, let the breadth be x m.

2) So, according to the problem, the length is more than twice its breadth,

breadth = (2x + 1) m

3) The area of the plot is 528 m², so we have,

x(2x + 1) = 528

2x² + x = 528

2x² + x – 528 = 0 ------------ equation 1

4) So, 2x² + x – 528 = 0 is the required quadratic equation.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

1) Let the first integer be x.

2) So, the next integer will be (x + 1).

3) The product of the integers is 306, so we have,

x(x + 1) = 306

x² + x = 306

x² + x – 306 = 0 ------------ equation 1

4) So, x² + x – 306 = 0 is the required quadratic equation.

(iii) 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.

1) Here, the age of Rohan's mother is depending on him, so let Rohan's age be x.

2) So, according to the problem, the age of Rohan's mother will be (x + 26).

3) 3 years later, Rohan's age will be (x + 3), and his mother's age will be (x + 29). 

4) According to the problem,

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

x(x + 29) + 3(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 = 360
x² + 32x + 87 – 360 = 0
x² + 32x – 273 = 0 ------------ equation 1

4) So, x² + 32x – 273 = 0 is the required quadratic equation.

(iv) 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.

1) Let the uniform speed of a train be x km/h.

2) To cover 480 km, the time taken will be (480/x) hrs.

3) If the speed is reduced by 8 km/h, then time will be increased by 3 hrs, so 

new speed = (x - 8) km/h, and new time = [(480/x) + 3] hrs.

4) According to the condition,

(x - 8)[(480/x) + 3] = 480

(x - 8)[(480 + 3x)/x] = 480

x(x - 8) = (480 + 3x)

x² - 8x = 480 + 3x

x² - 8x - 480 - 3x = 0
x² - 8x - 3x - 480 = 0

x² - 11x - 480 = 0 ------------ equation 1

5) So, x² - 11x - 480 = 0 is the required quadratic equation.

Click here for ⇨ NCERT-10-4-Quadratic Equations - Ex-4.2

ANIL SATPUTE