## Quadratic Equation Class 10 Ncert Solutions PDF Summary

Hello friends, today we are going to give the **Quadratic Equation Class 10 NCERT Solutions PDF **for your assistance. if you are searching Quadratic Equation Class 10 Ncert Solutions in PDF format but are unable to find it anywhere don’t worry you are reached at the right website. Also, you can download Quadratic Equation Class 10 Ncert Solutions PDF by clicking on the link given below. Via this post, learners can effortlessly get NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations. It includes all the solutions to the problems provided in Class 10 Maths NCERT textbook for second term exam preparations.

NCERT Solutions for Class 10 is a thorough and step-by-step guide to all the questions of the students which can be very helpful for the exam practice of all the students. Those learners who want to good scores in the assessments then can read this editorial properly. These NCERT Solutions assist the students in learning of the students.

### Quadratic Equation Class 10 Ncert Solutions PDF – Detailed Table of Chapter 4 Notes

1. |
Board |
CBSE |

2. |
Textbook |
NCERT |

3. |
Class |
Class 10 |

4. |
Subject |
Notes |

5. |
Chapter |
Chapter 4 |

6. |
Chapter Name |
Quadratic Equations |

7. |
Category |
CBSE Revision Notes |

### Quadratic Equations class 10 PDF

**Exercise 4.1 Page: 7**

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

(i) (x + 1)^{2} = 2(x – 3)

(ii) x^{2} – 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}

**Solutions:**

(i) Given,

(x + 1)^{2} = 2(x – 3)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

⇒ x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 7 = 0

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

Therefore, the given equation is quadratic equation.

(ii) Given, x^{2} – 2x = (–2) (3 – x)

⇒ x^{2 }–^{ }2x = -6 + 2x

⇒ x^{2 }– 4x + 6 = 0

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

Therefore, the given equation is quadratic equation.

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

By multiplication

⇒ x^{2 }– x – 2 = x^{2 }+ 2x – 3

⇒ 3x – 1 = 0

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

(iv) Given, (x – 3)(2x +1) = x(x + 5)

By multiplication

⇒ 2x^{2 }– 5x – 3 = x^{2 }+ 5x

⇒ x^{2 }– 10x – 3 = 0

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

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

By multiplication

⇒ 2x^{2 }– 7x + 3 = x^{2 }+ 4x – 5

⇒ x^{2 }– 11x + 8 = 0

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

Therefore, the given equation is quadratic equation.

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

By using the formula for (a-b)^{2}=a^{2}-2ab+b^{2}

⇒ x^{2} + 3x + 1 = x^{2} + 4 – 4x

⇒ 7x – 3 = 0

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

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

By using the formula for (a+b)^{3 }= a^{3}+b^{3}+3ab(a+b)

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

⇒ x^{3} + 14x – 6x^{2} – 8 = 0

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

Therefore, the given equation is not a quadratic equation.

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

By using the formula for (a-b)^{3 }= a^{3}-b^{3}-3ab(a-b)

⇒ x^{3} – 4x^{2} – x + 1 = x^{3} – 8 – 6x^{2 } + 12x

⇒ 2x^{2} – 13x + 9 = 0

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

Therefore, the given equation is quadratic.

**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 meters) 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**

**Solutions:**

(i) Let us consider,

The breadth of the rectangular plot = x m

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

As we know,

Area of rectangle = length × breadth = 528 m^{2}

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x^{2} + x =528

⇒ 2x^{2} + x – 528 = 0

Therefore, the length and breadth of the plot, satisfy the quadratic equation, 2x^{2} + x – 528 = 0, which is the required representation of the problem mathematically.

(ii) Let us consider,

The first integer number = x

Thus, the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306

⇒ x^{2 }+ x = 306

⇒ x^{2 }+ x – 306 = 0

Therefore, the two integers x and x+1 satisfy the quadratic equation, x^{2 }+ x – 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider,

Age of Rohan’s = x years

Therefore, as per the given question,

Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(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, which is the required representation of the problem mathematically.

(iv) Let us consider,

The speed of train = *x* km/h

And

Time taken to travel 480 km = 480/x km/hr

As per second condition, the speed of train = (*x* – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = (480/x)+3 km/h

As we know,

Speed × Time = Distance

Therefore,

(*x* – 8)(480/*x* )+ 3 = 480

⇒ 480 + 3*x* – 3840/*x* – 24 = 480

⇒ 3*x* – 3840/*x* = 24

⇒ *x*^{2 }– 8*x* – 1280 = 0

Therefore, the speed of the train, satisfies the quadratic equation, *x*^{2 }– 8*x* – 1280 = 0, which is the required representation of the problem mathematically.

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