Class10 NCRT Quadratic Equations Exercise – 4.1 pdf || UP Board

Everyday

August 06, 2021 August 06, 2021

Quadratic Equations
Exercise – 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x – 3)

on simplifying the equation,

(x + 1)² = 2(x – 3)

x² + 2x + 1 = 2x – 6

x² + 7 = 0

Or x² + 0x + 7 = 0

This equation is of the type ax² + bx + c = 0.

Hence, the given equation is a quadratic equation.

(ii) x² – 2x = (-2)(3 – x)

on simplifying the equation,

x² – 2x = (-2)(3 - x)

x² – 2x = -6 + 2x

x² – 4x + 6 = 0

Or x² – 4x + 6 = 0

This equation is of the type ax² + bx + c = 0.

Hence, the given equation is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

on simplifying the equation,

(x – 2)(x + 1) = (x – 1)(x + 3)

x² – 2x + x – 2 = x² – x + 3x – 3

-3x + 1 = 0

Or 3x – 1 = 0

This equation is not of the type ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)

on simplifying the equation,

2x² – 6x + x – 3 = x² + 5x

x² – 10x – 3 = 0

Or x² +(-10)x + (-3) = 0

This equation is of the type ax² + bx + c = 0.

Hence, the given equation is a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

on simplifying the equation,

2x² – x – 6x + 3 = x² + 5x – x – 5

x² – 11x + 8 = 0

This equation is of the type ax² + bx + c = 0.

Hence, the given equation is a quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

on simplifying the equation,

x² + 3x + 1 = x² – 4x + 4

7x – 3 =0

This equation is not of the type ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(vii) (x + 2)³ = 2x(x² – 1)

on simplifying the equation,

x³ + 6x² + 12x + 8 = 2x³ – 2x

-x³ + 6x² + 14x + 8 = 0

अर्थात x³ – 6x² – 14x – 8 = 0

This equation is not of the type ax2 + bx + c = 0.

Hence, the given equation is not a quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

on simplifying the equation,

x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8

2x² – 13x + 9 = 0

2x² + (– 13)x + 9 = 0

This equation is of the type ax² + bx + c = 0.

Hence, the given equation is a quadratic equation.

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.

Solution:

Let the breadth of the plot = x m

Hence, length of plot = 2x + 1 m

Hence, area = x(2x + 1) m²

According to Question,

x(2x + 1) = 528

2x² + x = 528

2x² + x – 528 = 0

Hence, the quadratic equation 2x² + x – 528 = 0 represents the length and breadth of the plot.

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

Let first integer = x

Therefore, second integer = x + 1

So, product = x(x + 1)

According to Question,

x(x + 1) = 306

x² + x = 306

x² + x – 306 = 0

Hence, the quadratic equation x² + x – 306 = 0 represents the product of two consecutive positive 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 Rohon’s present age.

Let Rohan's age = x years

Therefore, age of Rohan's mother = x + 26 years

after three years,

Let Rohan's age = x + 3 years

Therefore, age of Rohan's mother = x + 29 years

Therefore, product of ages = (x + 3)(x + 29) years

According to Question,

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

x² + 3x + 29x + 87 = 360

x² + 32x – 273 = 0

Hence, the quadratic equation x² + 32x – 273 = 0 represents the Rohon’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.

Let speed of train = x km/h

Total distance = 480 km

Hence, time taken by the train = 480/x hours

If its speed had been 8 km/m less, then time taken = 480/(x – 8) hours

According to Question,