# NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

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## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)2 = 2(x – 3)
(ii) x – 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
Solution:
(i) (x+ 1)2 = 2(x – 3)
⇒ x2 + 2x +1 = 2x – 6
⇒ x2 + 2x – 2x+1 + 6 = 0
⇒ x2 + 7 = 0
Now, it is in the form of ax² + bx + c = 0, where b = 0.
Therefore, the given equation is a quadratic equation.

(ii) x2 – 2x = (- 2) (3 – x)
⇒ x2 – 2x = – 6 + 2x
⇒ x2 – 4x + 6 = 0
Which is of the form
ax2 + bx + c = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(iii) We have,
(x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x2 + x – 2x – 2 = x² + 3x – x – 3
⇒ x2 + x – 2x – 2 = x2 – 3x + x + 3 = 0
⇒ – 3x + 1 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(iv) We have
(x-3) (2x+ 1) = x (x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 + x – 6x – 3 – x² – 5x = 0
⇒ x2 – 10x – 3 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(v) We have
(2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x2 – 6x-x + 3 = x2 -x + 5x – 5
2x2 – 6x-x + 3 = x2 + x – 5x + 5 = 0
⇒ x2 – 11x + 8 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(vi) We have
x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ x2 + 3x + 1 = x2– 4 + 4c = 0
⇒ 7x – 3 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(vii) We have
(x + 2)3 = 2x (x2 – 1)
⇒ x3 + 8 + 3.x.2 (x + 2) = 2x3 – 2x
⇒ x3 + 8 + 6x2 + 12x = 2x3 – 2x
⇒ x3 – 6x2 – 14x – 8 = 0
It is not in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

(viii) We have
x3 – 4x2 – x+1 = (x-2)3
⇒ x3 – 4x2 – x + 1 = x3-8 + 3x(-2)(x – 2)
⇒ x3 – 4x2 -x + 1 = x3 – 6x2 + 12x – 8
⇒ 2x2 – 13x + 9 = 0
Now, it is in the form of ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.

Question 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 breadth of the rectangular plot = x m
Then, length of the plot = (2x + 1)m
Area of a rectangular plot = l x b ,
⇒ 528 (2x + 1)x
⇒ 528 = 2x2 +x
⇒ 2x2 + x – 528 = 0
Which is the required quadratic equation.
Therefore, area of rectangle, satisfies the quadratic equation 2x² + x – 528 = 0, where x is breadth (in metres) of the plot.

(ii) Let first consecutive positive integer = x
∴ Second consecutive positive integer = x + 1
According to question,
x (x + 1) = 306
or x² + x = 306
or x² + x – 306 = 0
Therefore, the two consecutive positive integers whose product is 306, satisfies the quadratic equation.
x² + x – 306 = 0, where x is the smallest integer.

(iii) Let the present age of Rohan = x years
∴ Rohan’s mother’s present age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
After 3 years, Rohan’s mother’s age = (x + 26 + 3) years
According to question,
(x + 3) (x + 29) = 360
⇒ x2 + 29x + 3x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Therefore, product or Rohan’s and his mother’s age three years now satisfies the quadratic equation
x² + 32x – 273 = 0, where x (in years) is the present age of Rohan.

(iv) Case I :
Let the uniform speed of the train = u km/h
Total distance covered by the train = 480 km.
distance
speed
Case: II
Speed of train = (u – 8) km/h
and total distance covered by the train = 480 km
∴ Time taken = $$\frac { distance }{ speed }$$ = $$\frac { 480 }{ u }$$h
Case: II
Speed of train = (u – 8) km/h
and total distance covered by the train = 480 km
∴ Time taken = $$\frac { distance }{ speed }$$ = $$\frac { 480 }{ u – 8 }$$h
According to question,

Therefore, speed of train satisfies the quadratic equation 3u2 – 24K – 1280, where K (in km/h) is the speed of the train.

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