Quadratic Equation
Mathematics Class Tenth
NCERT Exercise 4.1
Question(1). Check whether the following are quadratic equations:
Question(1)(i)(x + 1)2 = 2 (x – 3)
Solution:
LHS = (x + 1)2
= x2 + 2x + 1
[∵(a + b)2 = ax2 + 2ab + b2]
Now, RHS = 2 (x – 3)
= 2x – 6
:. x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – (2x – 6) = 0
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 7 = 0
∵ given equation is in the form of ax2 + bx + C = 0, thus it is a quadratic equation
Question(1)(ii)x2 – 2x = ( – 2)(3 – x)
Solution:
Given, x2 – 2x = ( – 2) (3 – x)
⇒ x2 – 2x = – 6 + x
⇒ x2 – 2x – ( – 6 + x) = 0
⇒ x2 – 2x – 6 – x = 0
⇒ x – 3x + 6 = 0
∵ given equation is in the form of ax2 + bx + C = 0, thus it is a quadratic equation
Question(1)(iii)(x – 2)(x + 1)=(x – 1 (x + 3)
Solution:
Given,
(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
= x2 – x – 2 – x2 – 2x + 3 = 0
= – x – 2 – 2x + 3 = 0
⇒ – 4x + 1 = 0
Here the given equation is not in the form of ax2 + bx + C = 0, thus it is not a quadratic equation
Question(1)(iv)(x – 3)(2x + 1) = x (x + 5)
Solution:
Given, (x – 3)(2x + 1) = x (x + 5)
⇒ 2x2 + x – 6x + 3 =x2 + 5x
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – (x2 + 5x) = 0
⇒ 2x2 – 5x + 3 – x2 – 5x = 0
⇒ x2 – 5x – 3 – 5x = 0
⇒ x2 – 10x + 3 = 0
∵ given equation is in the form of ax2 + bx + C = 0, thus it is a quadratic equation
Question(1)(v)(2x – 1)(x – 3) = (x + 5)(x – 1)
Solution:
Given, (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
= 2x2 – 7x + 3 – x2 – 4x + 5 = 0
⇒ 2x2 – x2 – 7x – 4x + 5 + 3 = 0
⇒ x2 – 11x + 8 = 0
∵ given equation is in the form of ax2 + bx + C = 0, thus it is a quadratic equation
Question(1)(vi) x2 + 3x + 1 = (x – 2)2
Solution:
Given, x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
[∵ (a – b)2 = a2 – 2ab + b2]
⇒ x2 + 3x + 1 – (x2 – 4x + 4)
⇒ x2 + 3x + 1 – x2 + 4x – 4 = 0
⇒ x2 – x2 + 3x + 4x + 1 – 4 = 0
⇒ 7x – 3 = 0
Here the given equation is not in the form of ax2 + bx + C = 0, thus it is not a quadratic equation
Question(1)(vii)(x – 2)3 = 2x(x2 – 1)
Solution:
Given, (x – 2)3 = 2x(x2 – 1)
⇒ x3 + (3 × x2 × 2) + 3x (2)2 + 8 = 2x3 – 2x
[∵ (a+3)3 = a3 + 3a2 b + 3ab2 + b3]
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – (2x3 – 2x) = 0
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ (x3 – 2x3) + 6x2 + 12x + 2x + 8 = 0
⇒ – x3 + 6x2 + 14x + 8 = 0
Here the given equation is not in the form of ax2 + bx + C = 0, thus it is not a quadratic equation
Question(1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution:
Given,x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = x3 – (3x2 × 2) + (3x × 22) – 23
[∵ (a – 3)3 = a3 – 3a2 b + 3ab2 – b3]
⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ x3 – 4x2 – x + 1 – (x3 – 6x2 + 12x – 8) = 0
⇒ x3 – 4x2 – x + 1 – x3 + 6x2 – 12 x + 8 = 0
⇒ x3 – x3 – 4x2 + 6x2 – x – 12x + 1 + 8 = 0
⇒ 2x2 – 13x + 9 = 0
∵ given equation is in the form of ax2 + bx + C = 0, thus it is a quadratic equation
Question (2) Represent the following situation in the form of quadratic equations:
Qoestion (2)(i) The area of a rectangular plot is 528 m2. 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:
Given, Area = 528m2
And, length = (2 x breadth) + 1
Let, breadth = x
∴ Lenght = 2x + 1
∵ Area = Lenght x Breadth
:. 528 = (2x + 1) × x
⇒ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0
This equation is in the form of quadratic equation ax2 + bx + C = 0
Question(2)(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let, first integer = x
∴ Next consecutive integer = x + 1
Now, given, product of two consecutive integer = 306
∴ x × (x + 1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
This is in the form of a quadratic equation
Question(2) (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 age.
Solution:
Let, present age of Rohan = x
∴ present age of Rohan's mother = x + 26
Three years from now,
Rohan's age = x + 3
∴ Rohan's mother age = x + 26 + 3 = x + 29
Given, product of their age = 360
∴ (x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
= x2 + 32x – 273 = 0
This is in the form of a quadratic equation
Question(2)(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/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, speed of the train = x km/h
∴ reduced speed of train = (x – 8) km/h
Let, time taken at normal speed = t hour
∴ time taken at reduced speed = t + 3 hour
We know that, time = distance/speed
∴ t = 480/x – – – (i)
And t + 3 = 480/x – 8
⇒ t = 480/x – 8 – 3 – – – (ii)
Now from equations (i) and (ii)
480/x = 480/x – 8 – 3
⇒ 480/x = 480 – 3(x – 8)/x – 8
⇒ 480/x = 480 – 3x + 24/x – 8
⇒ 480/x = 504 – 3x/x – 8
After doing cross multiplication, we get
⇒ 480(x – 8) = x (504 – 3x)
⇒ 480x – 3840 = 504x – 3x2 = 0
⇒ 480x – 3840 – (504x – 3x2) = 0
⇒ 480x – 3840 – 504x + 3x2 = 0
⇒ 3x2 + 480x – 504x – 3840 = 0
⇒ 3x2 – 24x – 3840 = 0
After dividing both sides by 3, we get
3 x2/3 – 24 x/3 – 3840/3 = 0
⇒ x2 – 8x – 1280 = 0
⇒ x2 + 8( – x) + 1280( – 1) = 0
This is in the form of a quadratic equation
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