Quadratic Equations

Important Questions

Here are some important Class 10 Mathematics questions for Chapter 4, Quadratic Equations, thoughtfully curated to aid students in their preparation for the CBSE Class 10 Mathematics Examination 2024-25. By practicing a diverse range of question types, students can clarify doubts and enhance their problem-solving skills, leading to improved performance in the Quadratic Equations chapter.

Table of Contents

In Chapter 4 of Class 10 Mathematics,Quadratic Equations we will explore the standard form of a quadratic equation.we will delve into the methods of solving quadratic equations, both by factorization and by utilizing the quadratic formula. And we will discuss the Situational problems based on quadratic equations related to day to day activities to be incorporated.

A quadratic equation is a second-degree polynomial equation of the form ax^2 + bx + c = 0, where ‘x’ is the variable, and ‘a’, ‘b’, and ‘c’ are constants with ‘a’ not equal to zero. The highest power of ‘x’ in a quadratic equation is 2, making it a second-degree equation.
The standard form of a quadratic equation is represented as: ax^2 + bx + c = 0

**Ans.** (a)

**Explanation:**

Given, 2x^2 + x - 1 = 0 \\
⇒ 2x^2 + 2x - x - 1 = 0 \\
⇒ 2x (x + 1) - 1 (x + 1) = 0 \\
⇒ (x + 1)(2x - 1) = 0 \\
⇒ \text{x = -1 or x =}{\Large\frac{1}{2}} \\
\text{Hence roots of equation are -1 and }{\Large\frac{1}{2}}

\textbf{(b)}\sqrt{6}

(c) 5

\textbf{(d)} 6\sqrt{6}

**Ans.** (c)

**Explanation:**

For a given equation to have equal roots D = 0, D = b^2 - 4ac

i.e.Here, b = p, a = 2, c = 3

⇒ p^2 - 4 × 2 × 3 = 0 \\
⇒ p^2 - 24 = 0 \\
⇒ \text{p = }\sqrt{24} = 2\sqrt{6} \\
\text{Hence, for} \space p = 2\sqrt{6}\\
\text{The given equation has equal roots.}

**Ans.**

k \geq \space \text{Given quadratic equation is} \space Kx^2 + 11x - 4 = 0 \space \text{On comparing with} \space ax^2 + bx + c = 0, \\
\text{we get} \space a = k, b = 11, c = - 4 \\
∵ \text{It has real roots. } \\
⇒ ∴ D\geq 0 \\
⇒ b - 4ac\geq 0 \\
⇒ (11) - 4 × k × (- 4)\geq 0 \\
⇒ 121 + 16k \geq 0 \\
\text{k}\geq {\Large \frac{-121}{16}}

**Ans.** 12 years

**Explanation:**

Let the sister’s age be x. Thus, the girl’s age is 2x. Four years hence,

Sister’s age = (x + 4) and Girl’s age = (2x + 4)

Now (x + 4) (2x + 4) = 160

⇒ 2x^2 + 8x + 4x + 16 = 160 \\
⇒ 2x^2 + 12x + 16 = 160 \\
⇒ 2x^2 + 12x – 144 = 0 \\
⇒ x^2 + 6x – 72 = 0 \\
⇒ x^2 + 12x – 6x – 72 = 0 \\
⇒ x(x + 12) – 6(x + 12) = 0 \\
⇒ (x + 12) (x – 6) = 0 \\
⇒ x = 6 \space or \space – 12 \\
⇒ x = 6

(as age is always positive)Thus, the sister’s age is 6 years and the girl’s age is 2 × 6 = 12 years.

(i) coincident lines

(ii) intersecting lines

**Ans.** The two numbers are 7 and 2.

**Explanation:**

Let the natural numbers be x and y respectively.

Now, x – y = 5 …(i)

Also, {\Large \frac{1}{y}}-{\Large \frac{1}{x}}={\Large \frac{5}{14}} \\
⇒ 14x – 14y = 5xy \\
⇒ 14(x – y) = 5xy \\
⇒ 14 × 5 = 5xy [From (i)] \\
⇒ xy = 14 \\
⇒ x = {\Large \frac{14}{y}} \\
\text{Substituting} \\
x = {\Large \frac{14}{y}} \\ \text{Substituting} \\
{\Large \frac{14}{y}}- y = 5 \\
⇒ y -{\Large \frac{14}{y}} + 5 = 0 \\
⇒ y^2 + 5y – 14 = 0 \\
⇒ y^2 + 7y – 2y – 14 = 0 \\
⇒ y( y + 7) – 2( y + 7) = 0 \\
⇒ (y + 7) ( y – 2) = 0 \\
⇒ y = 2, – 7 \\
\text{Since natural numbers cannot be –ve, so y = 2.}\\
\text{Thus, from equation (i),x – y = 5 } \\
⇒ x – 2 = 5 \\
⇒ x = 7 \\
\text{So, the two numbers are 7 and 2.}

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

If you want to improve your understanding of the concepts in this chapter, you can visit oswal.io. They have a wide range of questions that will help you practice and reinforce what you’ve learned. By solving these questions, you can strengthen your knowledge and become better at solving problems.

Ans: A quadratic equation can have two solutions, one solution (double root), or no real solutions (complex roots), depending on the value of the discriminant (b^2 – 4ac).

Ans: Yes, a quadratic equation can have complex roots (imaginary roots) if the discriminant is negative.

Ans: To solve a quadratic equation by factorization, you need to express it as a product of two binomials and set each factor equal to zero to find the solutions.

Ans: You can determine the number of solutions by looking at the value of the discriminant:

- If the discriminant is greater than zero, the equation has two distinct real roots.
- If the discriminant is zero, the equation has one real root (double root).
- If the discriminant is negative, the equation has no real solutions (complex roots).

Ans: A quadratic equation is a second-degree polynomial equation (ax^2 + bx + c = 0), while a linear equation is a first-degree polynomial equation (ax + b = 0). The highest power of the variable ‘x’ in a quadratic equation is 2, while in a linear equation, it is 1.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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