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NCERT solutions for class 10 Maths chapter 4 Quadratic Equations (Exercise 4.1)

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NCERT solutions for class 10 Maths


Chapter 4


Quadratic Equations


Exercise 4.1


Ex 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:

Formula used – (a+b)= a2+2ab+b2

(i) (x + 1)2 = 2(x – 3)

⇒ x2 + 2x + 1 = 2x – 6

⇒ x2 + 7 = 0

The equation is in the form of ax2 + bx + c = 0.

Hence, the equation is quadratic equation.

(ii) x2 – 2x = (–2) (3 – x)

⇒ x 2x = -6 + 2x

⇒ x– 4x + 6 = 0

The equation is in the form of ax2 + bx + c = 0.

Hence, the equation is quadratic equation.

(iii)  (x – 2)(x + 1) = (x – 1)(x + 3)

⇒ x– x – 2 = x+ 2x – 3

⇒ 3x – 1 = 0

The equation is not in the form of ax2 + bx + c = 0.

Hence, the equation is not a quadratic equation.

(iv) (x – 3)(2x +1) = x(x + 5)

⇒ 2x– 5x – 3 = x+ 5x

⇒  x– 10x – 3 = 0

The equation is in the form of ax2 + bx + c = 0.

Hence, the given equation is quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x– 7x + 3 = x+ 4x – 5

⇒ x– 11x + 8 = 0

The equation is in the form of ax2 + bx + c = 0.

Hence, the given equation is quadratic equation.

(vi)  x2 + 3x + 1 = (x – 2)2

⇒ x2 + 3x + 1 = x2 + 4 – 4x

⇒ 7x – 3 = 0

The equation is not in the form of ax2 + bx + c = 0.

Hence, the equation is not a quadratic equation.

(vii) (x + 2)3 = 2x(x2 – 1)

⇒ x3 + 8 + x2 + 12x = 2x3 – 2x

⇒ x3 + 14x – 6x2 – 8 = 0

The equation is not in the form of ax2 + bx + c = 0.

Hnece, the equation is not a quadratic equation.

(viii) x3 – 4x2 – x + 1 = (x – 2)3

⇒  x3 – 4x2 – x + 1 = x3 – 8 – 6x + 12x

⇒ 2x2 – 13x + 9 = 0

The equation is in the form of ax2 + bx + c = 0.

Hence, the equation is quadratic equation.


Ex 4.1 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

The length of the plot = (2x + 1) m.

Area of rectangle = 528 m2   

(2x + 1) × x = 528

⇒ 2x2 + x =528

⇒ 2x2 + x – 528 = 0

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

(ii) Let the first integer number = x

∴ Second  integer will be = x + 1

According to question = x × (x +1) = 306

⇒ x+ x = 306

⇒ x+ x – 306 = 0

Hence, the two integers satisfies the quadratic equation, x+ x – 306 = 0.

(iii) Let the age of Rohan = x  years

Therefore, Rohan’s mother’s age = x + 26

After 3 years, Rohan’s age = x + 3

And Rohan’s mother’s age = x + 26 + 3 = x + 29

The product of their ages = (x + 3)(x + 29) = 360

⇒ x2 + 29x + 3x + 87 = 360

⇒ x2 + 32x + 87 – 360 = 0

⇒ x2 + 32x – 273 = 0

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0.

(iv) Let the speed of train = x  km/h

Total distance = 480 km

Time taken = 480/x hr

If the speed had been 8 km/h less, then time taken = (x – 8) km/h

According to question

480 / x – 8 – 480 / x = 3

⇒ 480x – 480 (x – 8) / (x – 8) x = 3

⇒ 480x – 480 x + 3640 = 3(x – 8)x

⇒ 3640 = 3x² – 24x

⇒ 3x² – 24x – 3640 = 0

Therefore, the speed of the train, satisfies the quadratic equation, 3x– 8x – 1280 = 0.

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