HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Haryana State Board HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Question 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.
Solution :
(i) The given equation is (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0
⇒ 1.x² + 0.x + 7 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(ii) The given equation is :
x² – 2x = (- 2)(3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0
⇒ 1.x² + (- 4).x + 6 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(iii) The given equation is :
(x – 2)(x + 1) = (x – 1)(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
⇒ 0.x² + (- 3) x + 1 = 0
∵ a = 0
∴ It is not of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is not a quadratic equation.

(iv) The given equation is :
(x – 3) (2x + 1) = x (x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 = x² + 5x
⇒ 2x² – 5x – 3 = x² – 5x = 0
⇒ x² – 10x – 3 = 0
⇒ 1.x² + (- 10).x + (- 3) = 0
It is of the form ax² + bx + c = 0 where a ≠ 0.
Hence, the given equation is a quadratic equation.

(v) The given equation is :
(2x – 1)(x – 3) = (x + 5)(x- 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 = x² + 4x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
⇒ 1.x² + (- 11)x + 8 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(vi) The given equation is :
x² + 3x + 1= (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ x² + 3x + 1 – x² + 4x – 4 = 0
⇒ 7x – 3 = 0
⇒ 0.x² + 7.x + (- 3) = 0
∵ a = 0
∴ It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(vii) The given equation is :
(x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ x³+ 6x² + 12x + 8 – 2x³ + 2x = 0
⇒ – x³ + 6x² + 14x + 8=0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(viii) The given equation is :
x³ – 4x² – x +1= (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³– 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
⇒ 2x² – 13x + 9 = 0
⇒ 2.x² + (- 13).x + 9 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

Question 2.
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.
Solution :
(i) Let the breadth of plot he x m.
Then length = (2x + 1) m. According to question,
x(2x + 1) = 528
[∵ Area of rectangle = Length x Breadth]
⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Hence, the quadratic equation is 2x² + x – 528 = 0.

(ii) Let the two consecutive integers be x and (x + 1)
According to question, x(x+ 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Hence, the quadratic equation is x² + x – 306 = 0.

(iii) Let the Rohan’s present age be x years.
Then, his mother’s present age be (x + 26) years.
After three years, age of Rohan = (x + 3) years
After three years, age of his mother = x + 26 + 3
= (x + 29) years
According to question,
(x + 3)(x + 29) = 360
⇒ x² + 29x +3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Hence, the quadratic equation is x² + 32x – 273 = 0.

(iv) Let the speed of the train be x km/h
Time taken to travel 480 km = \(\frac{480}{x}\) hours
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
If speed is 8 km/h less then new speed = (x – 8) km/h
Time taken to travel 480 km = \(\frac{480}{x-8}\) hours
According to question,
\(\frac{480}{x-8}-\frac{480}{x}\) = 3
⇒ \(\frac{480 x-480(x-8)}{x(x-8)}\) = 3
⇒ 480x – 480x + 480 x 8 = 3x (x – 8)
⇒ 3840 = 3x² – 24x
⇒ 3x² – 24x – 3840 = 0
⇒ x² – 8x – 1280 = 0
Hence, the quadratic equation is x² – 8x – 1280 = 0.