Quadratic Equations – Exercise 4.1 – Class 10

1. Check whether the following are quadratic equations:

Solution:

(i) (x+1)2 = 2(x – 3)

x2 + 2x + 1 = 2x – 6

x2 + 7 = 0

It is of the form ax2 + bx + c.

Hence, the given equation is a quadratic equation.

(ii) x2 – 2x = (-2)(3 – x)

x2 – 2x = -6 + 2x

x2 – 4x + 6 = 0

It is of the form ax2 + bx + c.

Hence, the given equation is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

x2 + x – 2x + 2 = x2 + 3x – x – 3

x2 – x – 2 = x2 + 2x – 3

-3x + 1 = 0

3x – 1 = 0

It is not of the form ax2 + bx + c.

Hence, the given equation is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x (x + 5)

2x2 + x – 6x – 3  = x2 + 5x

x2 – 5x – 5x – 3 = 0

x2 – 10x – 3  = 0

It is of the form ax2 + bx + c.

Hence, the given equation is a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

2x2 – 6x – x + 3 = x2 – x + 5x – 5

x2 – 7x + 3 = 4x – 5

x2 – 11x + 8 = 0

It is of the form ax2 + bx + c.

Hence, the given equation is a quadratic equation

(vi) x2 + 3x + 1 = (x – 2)2

x2 – 3x + 1 = x2 – 4x + 4

7x – 3 = 0

It is of the form ax2 + bx + c.

Hence, the given equation is not a quadratic equation.

(vii) (x + 2)3 = 2x(x2 – 1)

x3 + 8 + 6x2 + 12x = 2x3 – 2x

x3 – 14x – 6x2 – 8  = 0

It is not of the form ax2 + bx + c.

Hence, the given equation is not a quadratic equation.

(vii) x3 – 4x2 – x + 1 = (x – 2)3

x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x

2x2 – 13x + 9 = 0

It is of the form ax2 + bx + c.

Hence, the given equation is a quadratic

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 the plot be ‘x’ m.

Hence, the length of the plot is (2x + 1) m.

Area of a rectangle = Length × Breadth

∴ 528 = x (2x + 1)

2x² + x – 528 =0

(ii) Let the consecutive integers be ‘x’ and ‘x + 1’.

It is given that their product is 306.

x(x + 1) = 306

2x² + x – 306 = 0

(iii)Let Rohan’s age be ‘x’.

Hence, his mother’s age = x + 26 3 years

hence, Rohan’s age = x + 3

Mother’s age = x + 26 + 3 = x + 29

It is given that the product of their ages after 3 years is 360.

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

x² + 29x + 3x + 87 = 360

x² + 32x – 273 = 0

(iv) Let the speed of train be x km/h.

Time taken to travel 480 km = 480/x hrs

In second condition, let the speed of train =(x – 8) km/h

It is also given that the train will take 3 hours to cover the same distance. Therefore, time taken to travel 480 km = [(480/x) + 3]km/h

Speed x time = distance

5 thoughts on “Quadratic Equations – Exercise 4.1 – Class 10”

1. (…I am so glad I am out of school, these past 45 years!)

In spite of the above thought, I remember now why I found university maths so fascinating! You’ve a very helpful blog for those still involved in the study of math.

Thank you for visiting and following The Old Fossil Writes!

Pearl

Liked by 1 person

1. Thank you so much 🙂 for your kind words 🙂 thank you again…
you have solid contents in your blog, which made me following you 🙂 keep posting

Liked by 1 person