**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1**

**1. Check whether the following are quadratic equation?**

**Solution:** An equation of the form ax^{2}+bx+c = 0, a ≠ 0 and a, b, c are real numbers is called quadratic equation.

**(i) (x+1) ^{2} = 2(x-3)**

x^{2}+2x+1 = 2x-6

x^{2}+2x+1 -2x+6 = 0

x^{2}+7 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 1≠ 0, b = 0, c = 7

Hence the given equation is quadratic equation.

**(ii) x ^{2}-2x = (-2)(3-x)**

x^{2}-2x = -6+2x

x^{2}-2x -2x+6 = 0

x^{2}-4x+6 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 1≠ 0, b = -4, c = 6

Hence the given equation is quadratic equation.

**(iii) (x-2)(x+1) = (x-1)(x+3)**

x^{2}-2x+x-2 = x^{2}-x+3x-3

x^{2}-2x+x-2 -x^{2}+x-3x+3= 0

-3x+1 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 0, b = -3, c = 1

Hence the given equation is not quadratic equation.

**(iv) (x-3)(2x+1) = x(x+5)**

x^{2}-6x+x-3 = x^{2}+5x

x^{2}-6x+x-3 -x^{2}-5x = 0

-10x-3 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 0, b = -10, c = -3

Hence the given equation is not quadratic equation.

**(v) (2x-1)(x-3) = (x+5)(x-1)**

2x^{2}-6x-x+3 = x^{2}+5x-x-5

2x^{2}-6x-x+3 -x^{2}-5x+x+5 = 0

x^{2}-11x+8 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 1 ≠ 0, b = -11, c = 8

Hence the given equation is quadratic equation.

**(vi) x ^{2}+3x+1 = (x-2)^{2}**

x^{2}+3x+1 = x^{2}-4x+4

x^{2}+3x+1 -x^{2}+4x-4= 0

7x-3 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 0, b = 7, c = -3

Hence the given equation is not quadratic equation.

**(vii) (x+2) ^{3} = 2x(x^{2}-1)**

x^{3}+6x^{2}+12x+8 = 2x^{3}-2x

x^{3}+6x^{2}+12x+8 -2x^{3}+2x = 0

x^{3 }– 6x^{2}-14x-8 = 0

It is not in the form of ax^{2}+bx+c = 0s

Hence the given equation is not a quadratic equation.

**(viii) x ^{3}-4x^{2}-x+1 = (x-2)^{3}**

x^{3}-4x^{2}-x+1 = x^{3}-6x^{2}+12x-8

x^{3}-4x^{2}-x+1 -x^{3}+6x^{2}-12x+8= 0

2x^{2}-13x+9 = 0

Comparing above equation with ax^{2}+bx+c = 0, we have a = 2 ≠ 0, b = -13, c = 9

Hence the given equation is quadratic equation.

**2. Represent the following situations in the form of quadratic equations:**

**(i) The area of rectangular plot is 528 m ^{2}. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot .**

**Solution:** Let the breadth of the plot is x meters

Then, the length is (2x + 1) meters.

Area = length × breadth

528 = x × (2x + 1)

528 = 2x^{2} + x

2x^{2} + x – 528 = 0

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

**Solution: **Let x and x+1 be two consecutive integers.

Then, x(x + 1) = 306

x^{2 }+ x – 306 = 0

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

**Solution:** Let x be the Rohan’s present age. Then, his mother’s present age is x+26

After 3 years theirs ages will be x+3 and x+26+3 = x+29 respectively.

Hence,

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

x^{2}+3x+29x+87 = 360

x^{2}+32x- 273 = 0

** (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:** Let the speed of the train be x km/h

Then it covers 480 km in hours.

If the speed is x-8 km/hr then it takes hours.

Hence, – = 3

= 3

480×8 = 3x(x-8)

x^{2}-8x = 160×8

x^{2}– 8x -1280 = 0

**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1**