NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Quadratic Equations |

Exercise |
Ex 4.1 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths 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) Given: (x+ 1)**

^{2}=2(x-3)⇒ x

^{2}+ 1 + 2x = 2x – 6

⇒ x

^{2}+ 1 + 2x – 2x + 1 = 0

⇒ x

^{2}+ 7 = 0

As the highest power of x is 2, so the given equation is

**quadratic.**

**(ii) Given: x ^{2} – 2x = (- 2) (3 -x)**

⇒ x

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

⇒ x

^{2}-4x + 6 = 0

As the highest power of x is 2, so the given equation is

**quadratic.**

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

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

⇒ x^{2} – x – 2 = x^{2} + 2x – 3

⇒ 3x – 1 = 0

As the highest power of x is 2, so the given equation is **quadratic.**

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

⇒ 2x^{2} – 6x + x – 3 = x^{2} + 5x

⇒ x^{2} – 10x -3 = 0

As the highest power of x is 2, so the given equation is **quadratic.**

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

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

⇒ x^{2} – 11x + 8 = 0

As the highest power of x is 2, so the given equation is **quadratic.**.

**(vi) Given: x ^{2} + 3x + 1 = (x – 2)^{2}**

⇒ x

^{2}+ 3x + 1 = x

^{2}+ 4 – 4x

⇒ 7x – 3 = 0

As the highest power of x is 1, so the given equation is

**not**

**quadratic.**

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

⇒ x

^{3}+ 8 + 6x

^{2}+ 12x = 2x

^{3}– 2x

⇒ x

^{3}– 6x

^{2}– 14x – 8 = 0

As the highest power of x is 1, so the given equation is

**not**

**quadratic.**

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

⇒ x

^{3}– 4x

^{2}-x + 1 = x

^{3}– 6x

^{2}+ 12x – 8

⇒ 2x

^{2}– 13x + 9 = 0

As the highest power of x is 1, so the given equation is

**quadratic.**

**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 will be = (2x + 1)m

∴ x(2x + 1) = 528

⇒ 2x^{2} + x – 528 = 0

This is required representation.

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

Then, x(x + 1) = 306

⇒ x^{2} + x – 306 = 0

This is required representation.

**(iii)** Let the present age of Rohan = x years

Then, his mother’s age will be = (x + 26) years

Three years from now:

Rohan’s age will be (x + 3) years.

His mother’s age will be = (x + 26 + 3)

= (x + 29) years

∴ (x + 3) (x + 29) = 360

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 32x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

This is required representation.

**
(iv)** Let speed of the train = x km/h

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