## 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) 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:**

**Formula used – (a+b) ^{2 }= a^{2}+2ab+b^{2}**

(i) (x + 1)^{2} = 2(x – 3)

⇒ x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 7 = 0

The equation is in the form of ax^{2} + bx + c = 0.

Hence, the equation is quadratic equation.

(ii) x^{2} – 2x = (–2) (3 – x)

⇒ x^{2 }–^{ }2x = -6 + 2x

⇒ x^{2 }– 4x + 6 = 0

The equation is in the form of ax^{2} + bx + c = 0.

Hence, the equation is quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

⇒ x^{2 }– x – 2 = x^{2 }+ 2x – 3

⇒ 3x – 1 = 0

The equation is not in the form of ax^{2} + bx + c = 0.

Hence, the equation is not a quadratic equation.

(iv) (x – 3)(2x +1) = x(x + 5)

⇒ 2x^{2 }– 5x – 3 = x^{2 }+ 5x

⇒ x^{2 }– 10x – 3 = 0

The equation is in the form of ax^{2} + bx + c = 0.

Hence, the given equation is quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x^{2 }– 7x + 3 = x^{2 }+ 4x – 5

⇒ x^{2 }– 11x + 8 = 0

The equation is in the form of ax^{2} + bx + c = 0.

Hence, the given equation is quadratic equation.

(vi) x^{2} + 3x + 1 = (x – 2)^{2}

⇒ x^{2} + 3x + 1 = x^{2} + 4 – 4x

⇒ 7x – 3 = 0

The equation is not in the form of ax^{2} + bx + c = 0.

Hence, the equation is not a quadratic equation.

(vii) (x + 2)^{3} = 2x(x^{2} – 1)

⇒ x^{3} + 8 + x^{2} + 12x = 2x^{3} – 2x

⇒ x^{3} + 14x – 6x^{2} – 8 = 0

The equation is not in the form of ax^{2} + bx + c = 0.

Hnece, the 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 – 6x^{2 } + 12x

⇒ 2x^{2} – 13x + 9 = 0

The equation is in the form of ax^{2} + 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 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

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

Area of rectangle = 528 m^{2 }

(2x + 1) × x = 528

⇒ 2x^{2} + x =528

⇒ 2x^{2} + x – 528 = 0

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

(ii) Let the first integer number = x

∴ Second integer will be = x + 1

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

⇒ x^{2 }+ x = 306

⇒ x^{2 }+ x – 306 = 0

Hence, the two integers satisfies the quadratic equation, x^{2 }+ 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

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 32x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x^{2} + 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, 3*x*^{2 }– 8*x* – 1280 = 0.