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 2023-24. 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.
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.
Answer: A quadratic equation is a second-degree polynomial equation of the form ax2 + 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: ax2 + bx + c = 0
Ans. (a)
Explanation:
Given, 2x2 + x - 1 = 0
⇒ 2x2 + 2x - x - 1 = 0
⇒ 2x (x + 1) - 1 (x + 1) = 0
⇒ (x + 1)(2x - 1) = 0
\(⇒ \text{x = -1 or x =}{\Large\frac{1}{2}}\)
\(\text{Hence roots of equation are -1 and }{\Large\frac{1}{2}}\)
Ans. (c)
Explanation:
For a given equation to have equal roots
D = 0
D = b2 - 4ac
i.e.Here, b = p, a = 2, c = 3
⇒ p2 - 4 × 2 × 3 = 0
⇒ p2 - 24 = 0
\(⇒ \text{p = }\sqrt{24} = 2\sqrt{6}\)
\(\text{Hence, for} p = 2\sqrt{6}\)
\(\text{The given equation has equal roots.}\)
Ans. \(k\geq {\Large \frac{-121}{16}}\)
Explanation:
Given quadratic equation is
Kx2 + 11x - 4 = 0
On comparing with ax2 + bx + c = 0, we get
a = k, b = 11, c = - 4∵ 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
⇒ 2x2 + 8x + 4x + 16 = 160
⇒ 2x2 + 12x + 16 = 160
⇒ 2x2 + 12x – 144 = 0
⇒ x2 + 6x – 72 = 0
⇒ x2 + 12x – 6x – 72 = 0
⇒ x(x + 12) – 6(x + 12) = 0
⇒ (x + 12) (x – 6) = 0
⇒ x = 6 or – 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.
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
⇒ \(x = {\Large \frac{14}{y}}\)
Substituting \(x = {\Large \frac{14}{y}}\) Substituting
\({\Large \frac{14}{y}}- y = 5\)
⇒ \(y -{\Large \frac{14}{y}}\) + 5 = 0
⇒ y2 + 5y – 14 = 0
⇒ y2 + 7y – 2y – 14 = 0
⇒ y( y + 7) – 2( y + 7) = 0
⇒ (y + 7) ( y – 2) = 0
⇒ y = 2, – 7
Since natural numbers cannot be –ve, so y = 2.
Thus, from equation (i),
x – y = 5
⇒ x – 2 = 5
⇒ x = 7
So, the two numbers are 7 and 2.
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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 (b2 - 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:
Ans: A quadratic equation is a second-degree polynomial equation (ax2 + 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.