## Quadratic Equations Class 10 Notes Maths Chapter 4

Quadratic equations are equations of the form ax^2 + bx + c = 0, where a, b and c are constants and x is a variable. In this chapter, we will learn how to solve quadratic equations using different methods.

- Introduction to Quadratic Equations:
- A quadratic equation is a polynomial equation of second degree, which means the highest power of the variable is 2.
- The standard form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants.
- The roots of a quadratic equation are the values of x that satisfy the equation.

- Methods to Solve Quadratic Equations:
- Factorization Method: In this method, we factorize the quadratic equation into two linear factors and solve for the values of x.
- Completing the Square Method: In this method, we convert the quadratic equation into the form (x + p)^2 = q and solve for the values of x.
- Quadratic Formula Method: In this method, we use the quadratic formula x = (-b ± √(b^2 – 4ac)) / 2a to solve for the values of x.

- Discriminant of a Quadratic Equation:
- The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 – 4ac.
- If D > 0, the quadratic equation has two real and distinct roots.
- If D = 0, the quadratic equation has one real and repeated root.
- If D < 0, the quadratic equation has two complex conjugate roots.

- Nature of Roots:
- The nature of the roots of a quadratic equation depends on the value of the discriminant.
- If the discriminant is positive, the roots are real and distinct.
- If the discriminant is zero, the roots are real and equal.
- If the discriminant is negative, the roots are complex conjugate.

- Sum and Product of Roots:
- If α and β are the roots of a quadratic equation ax^2 + bx + c = 0, then α + β = -b/a and αβ = c/a.

- Quadratic Equations in Real Life:
- Quadratic equations are used in various fields like physics, engineering, economics, and finance to model and solve problems.
- Some examples of real-life applications of quadratic equations are projectile motion, optimization, and profit maximization.

Overall, the chapter on Quadratic Equations is important for building a strong foundation in mathematics and problem-solving skills. The different methods and concepts covered in this chapter provide a useful toolkit for solving a wide range of problems in various fields.

**CBSE Class 10 Maths Important Questions Chapter 4 Quadratic Equations**

Important Questions:

- Solve the quadratic equation by factorization method: x^2 + 7x + 10 = 0.
- Solve the quadratic equation using the quadratic formula: 2x^2 + 5x + 3 = 0.
- Find the discriminant of the quadratic equation 4x^2 + 6x + 1 = 0.
- Determine the nature of the roots of the quadratic equation 2x^2 + 4x + 6 = 0.
- If one root of the quadratic equation ax^2 + bx + c = 0 is 2, find the other root in terms of a, b and c.
- A shopkeeper sells two types of toys A and B. She earns a profit of 20% on toy A and 15% on toy B. If she sells 100 toys and earns a profit of 18% on the whole, how many toys of each type did she sell? Use a quadratic equation to solve.
- A man standing on the top of a tower of height 10m sees a car moving towards the tower. If the angle of depression of the car changes from 30° to 45° in 5 seconds, find the speed of the car. Use a quadratic equation to solve.
- A rectangular field is to be fenced on three sides leaving a side of 20 meters uncovered. If the area of the field is 600 sq. meters, find the area enclosed by the fence. Use a quadratic equation to solve.
- A train covers a distance of 400km in 5 hours. Another train covers the same distance in 4 hours. Find the speed of both the trains. Use a quadratic equation to solve.
- A manufacturer produces nuts and bolts. The cost of producing a nut is Rs. 0.75 and that of a bolt is Rs. 1.25. He earns a profit of Rs. 200 per day. If he produces x nuts and y bolts per day, form a quadratic equation to represent this information.

**CBSE Class 10 Maths Important Questions Answers Chapter 4 Quadratic Equations**

- Find the quadratic equation whose roots are 2 and -5.

Solution: The quadratic equation with roots α and β is given by (x – α)(x – β) = 0 Therefore, the quadratic equation with roots 2 and -5 is given by (x – 2)(x + 5) = 0 Expanding the above equation, we get x^2 + 3x – 10 = 0 Hence, the required quadratic equation is x^2 + 3x – 10 = 0.

- Find the roots of the quadratic equation 4x^2 – 9x + 2 = 0.

Solution: We have the quadratic equation 4x^2 – 9x + 2 = 0. To find the roots, we use the quadratic formula. The quadratic formula is given by x = (-b ± √(b^2 – 4ac)) / 2a Here, a = 4, b = -9, and c = 2. Substituting these values in the formula, we get x = (-(-9) ± √((-9)^2 – 4(4)(2))) / 2(4) x = (9 ± √33) / 8 Therefore, the roots of the given quadratic equation are x = (9 + √33) / 8 and x = (9 – √33) / 8.

- Solve the quadratic equation 3x^2 + 2x – 1 = 0.

Solution: We have the quadratic equation 3x^2 + 2x – 1 = 0. To solve this equation, we can either use the quadratic formula or factorization method. Using the quadratic formula, we get x = (-b ± √(b^2 – 4ac)) / 2a Here, a = 3, b = 2, and c = -1. Substituting these values in the formula, we get x = (-2 ± √(2^2 – 4(3)(-1))) / 2(3) x = (-2 ± √22) / 6 Therefore, the roots of the given quadratic equation are x = (-2 + √22) / 6 and x = (-2 – √22) / 6. Alternatively, we can use the factorization method as follows: 3x^2 + 2x – 1 = 0 Multiplying both sides by 3, we get 9x^2 + 6x – 3 = 0 Dividing both sides by 3, we get 3x^2 + 2x – 1 = 0 Now, we can factorize the quadratic expression as (3x – 1)(x + 1) = 0 Therefore, the roots of the given quadratic equation are x = 1/3 and x = -1.

- Find the quadratic equation whose roots are reciprocal of each other.

Solution: Let the roots of the quadratic equation be α and β. We are given that the roots are reciprocal of each other, i.e., β = 1/α. Using the sum and product of roots formula, we have α + β = α + 1/α and αβ = α(1/α) = 1 We can now write the quadratic equation as x^2 – (α + 1/α)x + 1 = 0 Substituting β = 1/α in the above equation, we get

**CBSE Class 10 Maths Important Questions Answers MCQs Chapter 4 Quadratic Equations**

What is the discriminant of the quadratic equation ax^2 + bx + c = 0?

a) b^2 + ac

b) 2ac – b^2

c) b^2 – 4ac

d) 4ac – b^2

Answer: c) b^2 – 4ac

If the discriminant of a quadratic equation is negative, then what can we say about the roots of the equation?

a) The roots are real and unequal

b) The roots are real and equal

c) The roots are imaginary

d) The equation has no roots

Answer: c) The roots are imaginary

What is the formula to find the sum of the roots of a quadratic equation ax^2 + bx + c = 0?

a) -b/a

b) c/a

c) -b/2a

d) b^2 – 4ac

Answer: c) -b/2a

If the roots of a quadratic equation are -2 and 5, what is the equation?

a) x^2 – 3x – 10 = 0

b) x^2 + 3x – 10 = 0

c) x^2 – 3x + 10 = 0

d) x^2 + 3x + 10 = 0

Answer: a) x^2 – 3x – 10 = 0

What is the value of the discriminant of a quadratic equation with roots -3 and -4?

a) 7

b) 20

c) -7

d) -20

Answer: b) 20

## CBSE Class 10 Maths Notes Chapter 4 Quadratic Equations

A quadratic polynomial of the form ax² + bx + c, where a ≠ 0 and a, b, c are real numbers, is called a quadratic equation

when ax² + bx + c = 0.

Here a and b are the coefficients of x² and x respectively and ‘c’ is a constant term.

Any value is a solution of a quadratic equation if and only if it satisfies the quadratic equation.

**Quadratic formula:** The roots, i.e., α and β of a quadratic equation ax² + bx + c = 0 are given

by −b±D√2a or −b±b2−4ac√2a provided b² – 4ac ≥ 0.

Here, the value b² – 4ac is known as the discriminant and is generally denoted by D. ‘D’ helps us to determine the nature of roots for a given quadratic equation. Thus D = b² – 4ac.

**The rules are:**

- If D = 0 ⇒ The roots are Real and Equal.
- If D > 0 ⇒ The two roots are Real and Unequal.
- If D < 0 ⇒ No Real roots exist.

If α and β are the roots of the quadratic equation, then Quadratic equation is x² – (α + β) x + αβ = 0 Or x² – (sum of roots) x + product of roots = 0

where, Sum of roots (α + β) = −coefficientofxcoefficientofx2=−ba

Product of roots (α x β) = coefficienttermcoefficientofx2=ca

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