## Chapter 4 Quadratic Equations 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) x**

^{2}+ 3x + 1 = (x – 2)^{2}**(vii) (x + 2)**

^{3}= 2x(x^{2}– 1)**(viii) x**

^{3}-4x^{2}-x + 1 = (x-2)^{3}**Solution:**

**(i) (x+ 1)**

^{2}=2(x-3)⇒ x

^{2}+ 2x +1 = 2x – 6

⇒ x

^{2}+ 2x – 2x+1 + 6 = 0

⇒ x

^{2}+ 7 = 0

⇒ x

^{2}+ 0x + 1 = 0

Which is of the form

ax

^{2}+ bx + c = 0

Hence, the given equation is a quadratic equation.

**(ii) x ^{2} – 2x = (- 2) (3 -x)**

⇒ x

^{2}-2x = -6 + 2x

⇒ x

^{2}-4x + 6 = 0

Which is of the form

ax

^{2}+ bx + c = 0

Hence, the given equation is a quadratic equation.

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

⇒ x^{2} + x-2x-2 = x2 + 3x-x -3

⇒ x^{2} + x – 2x – 2 = x^{2} – 3x + x + 3 = 0

⇒ – 3x + 1 = 0 ⇒ 3x – 1 = 0

Since degree of equation is 1, hence, given equation is not a quadratic equation.

**(iv)** (x-3) (2x+ 1) = x (x + 5)

⇒ 2x^{2} + x – 6x – 3 = x^{2} + 5x

2x^{2} + x – 6x – 3-x2 – 5x = 0

⇒ x^{2} – 10x -3 = 0

Which is of the form

ax^{2} + bx + c 0

Hence, the given equation is a quadratic equation.

**(v) (2x-1)(x-3) = (x + 5)(x-1)**

⇒ 2x^{2} – 6x-x + 3 = x^{2} -x + 5x – 5

2x^{2} – 6x-x + 3 = x^{2} + x – 5x + 5 = 0

⇒ x^{2} – 11x + 8 = 0

Which is of the form

ax^{2} + bx + c = 0

Hence, the given equation is a quadratic equation.

**(vi) x ^{2} + 3x + 1 = (x-2)^{2}**

⇒ x

^{2}+ 3x + 1 = x

^{2}+ 4 – 4x

⇒ x

^{2}+ 3x + 1 = x

^{2}– 4 + 4c = 0

⇒ 7x – 3 = 0

Since degree of equation is 1, hence, the given equation is not a quadratic equation,

**(vii) (x + 2) ^{3} = 2x (x^{2} – 1)**

⇒ x

^{3}+ 8 + 3.x.2 (x + 2) = 2x

^{3}– 2x

⇒ x

^{3}+ 8 + 6x

^{2}+ 12x = 2x

^{3}– 2x

⇒ x

^{3}– 6x

^{2}– 14x – 8 = 0

Which is not of the form

ax

^{2}+ bx + c = 0

Hence, the given equation is not a quadratic equation.

**(viii) x ^{3} – 4x^{2} – x+1 = (x-2)^{3}**

⇒ x

^{3}– 4x

^{2}– x + 1 = x

^{3}-8 + 3x(-2)(x – 2)

⇒ x

^{3}– 4x

^{2}-x + 1 = x

^{3}– 6x

^{2}+ 12x – 8

⇒ 2x

^{2}– 13x + 9 = 0

Which is of the form

ax

^{2}+ bx + c = 0

Hence, 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 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)** 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 = 2x^{2} +x

⇒ 2x^{2} + x – 528 = 0

Which is the required quadratic equation.

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

Then, x(x+l) = 306

⇒ x^{2} +x-306 = 0

Which is the required quadratic equation.

**(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

⇒ x^{2} + 29x + 3x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

Which is the required quadratic equation. **(iv)** Let speed of the train = x km/h