The document Ex 4.1 NCERT Solutions- Quadratic Equations Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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**Ques 1: Check whether the following are quadratic equations:**

**(i) (x + 1) ^{2} = 2(x âˆ’ 3)(ii) x^{2} âˆ’ 2x = (âˆ’ 2) (3 âˆ’ x)(iii) (x âˆ’ 2) (x + 1) = (x âˆ’ 1) (x + 3)(iv) (x âˆ’ 3) (2x + 1) = x(x + 5)(v) (2x âˆ’ 1) (x âˆ’ 3) = (x + 5) (x âˆ’ 1)(vi) x^{2} + 3x + 1 = (x âˆ’ 2)^{2}(vii) (x + 2)^{3} = 2x(x^{2} âˆ’ 1)(viii) x^{3} âˆ’ 4x^{2} âˆ’ x + 1 = (x âˆ’ 2)**

**Sol: ****(i)** (x + 1)^{2} = 2(xâˆ’3)

We have:

(x + 1)^{2} = 2 (x âˆ’ 3)

â‡’ x^{2 }+ 2x + 1 = 2x âˆ’ 6

â‡’ x^{2} + 2x + 1 âˆ’ 2x + 6 = 0

â‡’ x^{2} + 7 = 0

Since x^{2} + 7 is a quadratic polynomial

âˆ´ (x + 1)^{2} = 2(x âˆ’ 3) is a quadratic equation.

**(ii)** x^{2}âˆ’2x = (âˆ’ 2) (3 âˆ’x)

We have:

x^{2} âˆ’ 2x = (âˆ’ 2) (3 âˆ’ x)

â‡’ x^{2} âˆ’ 2x = âˆ’ 6 + 2x

â‡’ x^{2} âˆ’ 2x âˆ’ 2x + 6 = 0

â‡’ x^{2} âˆ’ 4x + 6 = 0

Since x^{2 }âˆ’ 4x + 6 is a quadratic polynomial

âˆ´ x^{2} âˆ’ 2x = (âˆ’2) (3 âˆ’ x) is a quadratic equation.

**(iii) **(xâˆ’2) (x + 1) = (xâˆ’1) (x + 3)

We have:

(x âˆ’ 2) (x + 1) = (x âˆ’ 1) (x + 3)

â‡’ x^{2} âˆ’ x âˆ’ 2 = x^{2} + 2x âˆ’ 3

â‡’ x^{2} âˆ’ x âˆ’ 2 âˆ’ x^{2} âˆ’ 2x + 3 = 0

â‡’ âˆ’3x + 1 = 0

Since âˆ’3x + 1 is a linear polynomial

âˆ´ (x âˆ’ 2) (x + 1) = (x âˆ’ 1) (x + 3) is not quadratic equation.

**(iv)** (xâˆ’3) (2x + 1) = x(x + 5)

We have:

(x âˆ’ 3) (2x + 1) = x (x + 5)

â‡’ 2x^{2} + x âˆ’ 6x âˆ’3= x^{2 }+ 5x

â‡’ 2x^{2} âˆ’ 5x âˆ’ 3 âˆ’ x^{2} âˆ’ 5x =0

â‡’ x^{2} + 10x âˆ’ 3= 0

Since x^{2} + 10x âˆ’ 3 is a quadratic polynomial

âˆ´ (x âˆ’ 3) (2x + 1) = x(x + 5) is a quadratic equation.

**(v) **(2xâˆ’1) (xâˆ’3) = (x + 5) (xâˆ’1)

We have:

(2x âˆ’ 1) (x âˆ’ 3) = (x + 5) (x âˆ’ 1)

â‡’ 2x^{2} âˆ’ 6x âˆ’ x + 3 = x^{2} âˆ’ x + 5x âˆ’ 5

â‡’ 2x^{2} âˆ’ x^{2} âˆ’ 6x âˆ’ x + x âˆ’ 5x + 3 + 5 = 0

â‡’ x^{2} âˆ’ 11x + 8 = 0

Since x^{2} âˆ’ 11x + 8 is a quadratic polynomial

âˆ´ (2x âˆ’ 1) (x âˆ’ 3) = (x + 5) (x âˆ’ 1) is a quadratic equation.

**(vi)** x^{2} + 3x + 1 = (xâˆ’2)^{2}

We have:

x^{2} + 3x + 1 = (x âˆ’ 2)^{2}

â‡’ x^{2} + 3x + 1 = x^{2} âˆ’ 4x + 4

â‡’ x^{2} + 3x + 1 âˆ’ x^{2 }+ 4x âˆ’4= 0

â‡’ 7x âˆ’ 3 = 0

Since 7x âˆ’ 3 is a linear polynomial

âˆ´ x^{2} + 3x + 1 = (x âˆ’ 2)^{2 }is not a quadratic equation.

**(vii)** (x + 2)^{3} = 2x(x^{2}âˆ’1)

We have:

(x + 2)^{3} = 2x(x^{2} âˆ’ 1)

â‡’ x^{3} + 3x^{2}(2) + 3x(2)^{2} + (2)^{3} = 2x^{3} âˆ’ 2x

â‡’ x^{3} + 6x^{2} + 12x + 8 = 2x^{3} âˆ’ 2x

â‡’ x^{3} + 6x^{2} + 12x + 8 âˆ’ 2x^{3} + 2x = 0

â‡’âˆ’ x^{3} + 6x^{2} + 14x + 8 = 0

Since âˆ’x^{3} + 6x^{2} + 14x + 8 is a polynomial of degree 3

âˆ´ (x + 2^{)3} = 2x(x^{2} âˆ’ 1) is not a quadratic equation.

**(viii)** x^{3}âˆ’4x^{2}âˆ’x + 1 = (xâˆ’ 2)^{3}

We have:

x^{3} âˆ’ 4x^{2} âˆ’ x + 1 = (x âˆ’ 2)^{3}

â‡’ x^{3} âˆ’ 4x^{2} âˆ’ x + 1 = x^{3} + 3x^{2}(âˆ’ 2) + 3x(âˆ’ 2)^{2} + (âˆ’ 2)^{3}

â‡’ x^{3} âˆ’ 4x^{2} âˆ’ x + 1 = x^{3} âˆ’ 6x^{2} + 12x âˆ’ 8

â‡’ x^{3} âˆ’ 4x^{2} âˆ’ x âˆ’ 1 âˆ’ x^{3} + 6x^{2} âˆ’ 12x + 8 = 0

â‡’ 2x^{2} âˆ’ 13x + 9 = 0

Since 2x^{2} âˆ’ 13x + 9 is a quadratic polynomial

âˆ´ x^{3} âˆ’ 4x^{2} âˆ’ x + 1 = (x âˆ’ 2)^{3} is a quadratic equation.**Ques ****2:** **Represent the following situations in the form of quadratic equations:****(i) The area of a rectangular plot is 528 m ^{2}. 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) Rohanâ€™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 Rohanâ€™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 the breadth = x metres

âˆµ Length = 2 (Breadth) + 1

âˆ´ Length = (2x + 1) metres

Since Length Ã— Breadth = Area

âˆ´ (2x + 1) Ã— x = 528

â‡’ 2x^{2} + x = 528

â‡’ 2x^{2} + x âˆ’ 528 = 0

Thus, the required quadratic equation is 2x^{2} + xâˆ’ 528 = 0

**(ii)** Let the two consecutive numbers be x and (x + 1).

âˆµProduct of the numbers = 306

âˆ´ x (x + 1) = 306

â‡’ x^{2} + x = 306

â‡’ x^{2} + x âˆ’ 306 = 0

Thus, the required equdratic equation is x^{2} + x âˆ’ 306 = 0

**(iii) **Let the present age = x

âˆ´ Motherâ€™s age = (x + 26) years

After 3 years

His age = (x + 3) years

Motherâ€™s age = [(x + 26) + 3] years

=(x + 29) years

According to the condition,

â‡’ (x + 3) Ã— (x + 29) = 360

â‡’ x^{2} + 29x + 3x + 87 = 360

â‡’ x^{2} + 29x + 3x + 87 âˆ’ 360 = 0

â‡’ x^{2} + 32x âˆ’ 273 = 0

Thus, the required quadratic equation is

x^{2} + 32x âˆ’ 273 = 0

(iv) In first case,

Let the speed of the train = u km/hr

Distance covered = 480 km

Time taken = Distance Ã· Speed

In second case,

Speed = (u âˆ’ 8) km/hour

According to the condition,

â‡’ 480u âˆ’ 480(u âˆ’ 8) = 3u(u âˆ’ 8)

â‡’ 480u âˆ’ 480u + 3840 = 3u^{2} âˆ’ 24u

â‡’ 3840 âˆ’ 3u^{2} + 24u =0

â‡’ 1280 âˆ’ u^{2 }+ 8u =0

â‡’ âˆ’ 1280 + u^{2} âˆ’ 8u =0

â‡’ u^{2} âˆ’ 8u âˆ’ 1280 = 0

Thus, the required quadratic equation is u^{2} âˆ’ 8u â€“ 1280 = 0

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