Class 10 Quadratic Equations NCERT solutions Maths
Chapter 4: Quadratic Equations
TEXTBOOK’S EXERCISE 4.1
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
Sol. Standard form of quadratic equation is ax2 + bx + c = 0; a ≠ 0.
(i) (x + 1)2 = 2(x – 3)
⇒ x2 + 1 + 2x = 2(x – 3)
⇒ x2 + 1 + 2x = 2x – 6
⇒ x2 + 1 + 2x – 2x + 6 = 0 ⇒ x2 + 7 = 0
It is of the form ax2 + bx + c = 0,
Where a = 1, b = 0 and c = 7.
Hence, it is a quadratic equation.
(ii) x2 – 2x = (–2) (3 – x)
⇒ x2 – 2x = – 6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
It is of the form ax2 + bx + c = 0,
where a = 1, b = – 4 and c = 6. Hence, it is a quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x(x + 1) –2(x + 1) = x(x + 3) –1(x + 3)
⇒ x2 + x – 2x – 2 = x2 + 3x – x – 3
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ x2 + 2x – 3 – x2 + x + 2 = 0
⇒ 3x – 1 = 0
It is not of the form ax2 + bx + c = 0.
Hence, it is not a quadratic equation.
(iv) (x – 3)(2x + 1) = x(x + 5)
⇒ x(2x + 1) – 3(2x + 1) = x2 + 5x
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x = 0
⇒ x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0, where a = 1, b = –10 and c = –3. Hence, it is a quadratic equation.
Class 10 Quadratic Equations NCERT solutions Maths
It is of the form ax2 + bx + c = 0, where a = 1, b = –11 and c = 8.
Hence, it is a quadratic equation.
It is not of the form ax2 + bx + c = 0. Hence, the given equation is not a quadratic equation.
It is not of the form ax2 + bx + c = 0. Hence, the given equation is not a quadratic equation.
Class 10 Quadratic Equations NCERT solutions Maths
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.
Sol. (i) Let breadth of rectangular plot = x m
Then, length of the rectangular plot
= (2x + 1) m
So, area of rectangular plot
= [x (2x + 1)] m2 = (2x2 + x) m2
As per condition :
Hence, given problem in the form of quadratic equation is
(ii) Let two consecutive positive integers are x and x + 1.
Product of integers = x (x + 1) = x2 + x As per condition:
Hence, given problem in the form of quadratic equation is x2 + x – 306 = 0.
Hence, given problem in the form of quadratic equation is x2 + x – 306 = 0.
(iii) Let present age of Rohan = x years
Present age of Rohan’s mother = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
After 3 years, Rohan’s mother’s age
= (x + 26 + 3) years = (x + 29) years
So, product = (x + 3) (x + 29)
= x2 + 29x + 3x + 87
= x2 + 32x + 87
As per condition: x2 + 32x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Hence, given problem in the form of quadratic equation is x2 + 32x – 273 = 0.
(iv) Let speed of train = x km/hour
Distance covered by train = 480 km
Time taken by train = Distance/ Speed = 480 /x hour
If speed of train is decreased by 8 km/hour
Then, new speed of train = (x – 8) km/hour Then, Time taken by train = 480
As per condition:
⇒ 3840 = 3(x2 – 8x) ⇒ 3x2 – 24x = 3840
⇒ x2 – 8x = 1280
⇒ x2 – 8x – 1280 = 0
Hence, given problem in the form of quadratic equation is
x2 – 8x – 1280 = 0.