# Quadratic Equations Class 10 Notes Maths Chapter 4

• 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.

## How to Find the Solution of a Quadratic Equation by Factorisation?

For a quadratic equation, a real number α is called the root of a quadratic equation ax2+bx+c =0. Hence, we can write aα2 + bα + c = 0. So, x= α is the solution of a quadratic equation or the root of a quadratic equation. In other words, α satisfies the given quadratic equation.

Note: The zeros of the quadratic equation ax2+bx+c = 0 are the same as the root of the quadratic equation ax2+bx+c = 0.

Example: Solve the quadratic equation 2x2+x-300 = 0 by the factorisation method.

Sol:

Given quadratic equation: 2x2+x-300 = 0

By using factorisation, the quadratic equation 2x2+x -300 = 0 is written as:

2x2 – 24x+25x -300 = 0

2x(x-12) +25(x-12) =0

(i.e) (x-12)(2x+25) = 0

Therefore, x-12=0 and 2x+25 = 0

x-12 = 0

Therefore, x= 12.

Similarly, 2x+25 = 0

2x= -25

x =-25/2

x = -12.5.

Hence, the roots of the quadratic equation 2x2+x-300 = 0 are 12 and -12.5.

Question for Short Notes: Quadratic Equations
Try yourself:Which of the following is true about the roots of the quadratic equation ax2 + bx + c = 0?

• The roots, if a quadratic equation ax2 + bx + c = 0 are given by:

• 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.

Example: Find the roots of quadratic equation x2 - 7x + 10 = 0 using quadratic formula.

Sol:

Here, a = 1, b = -7 and c = 10. Then by quadratic formula:

x = [-(-7) ± √((-7)2 - 4(1)(10))] / (2(1))

= [ 7 ± √(49 - 40) ] / 2

= [ 7 ± √(9) ] / 2

= [ 7 ± 3 ] / 2

= (7 + 3) / 2, (7 - 3) / 2

= 10/2, 4/2

= 5, 2

Therefore, x = 2, x = 5.

## Nature of Roots of a Quadratic Equation

• 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.

Example: The roots of the quadratic equation 3x2-10x+3 = 0 are

a) real and equal

b) imaginary

c) real, unequal and rational

d) none of these

Sol:

Given equation 3x2-10x+3 = 0

Here discriminant, D = b2-4ac

= (-10)2 – 4×3×3

= 100 – 36

= 64

D is positive and a perfect square.

So the roots of the quadratic equation are real, unequal and rational.

Hence option (c) is the answer.

Question for Short Notes: Quadratic Equations
Try yourself:Given the quadratic equation ax2 + bx + c = 0, where a, b, and c are constants. What can be inferred about the roots of this equation if the discriminant (D = b2 - 4ac) is less than 0?

## Relationship between roots and Coefficients of Quadratic Equation

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) =
• Product of roots (α x β) =

Example: If α and β are the roots of the equation x2 - 4x + 2 = 0, find the value of

i) α2 + β2

ii) α2 - β2

iii) α3 - β3

iv)1/α  + 1/ β

Sol:

The given equation is x2 - 4x + 2 = 0 ...................... (i)

According to the problem, α and β are the roots of the equation (i)

Therefore,

(i) Now α2 + β2 = (α + β)2 - 2αβ = (4)2 – 2 x 2 = 16 – 4 = 12.

(ii) α2 - β2 = (α + β)( α - β)

Now (α - β)2 = (α + β)2 - 4αβ = (4)2 – 4 x 2 = 16 – 8 = 8

⇒ α - β = ± √8

⇒ α - β = ± 2√2

Therefore, α2 - β2 = (α + β)( α - β) = 4 x (± 2√2) = ± 8√2.

(iii) α3 + β3 = (α + β)3 - 3αβ(α + β) = (4)3 – 3 x 2 x 4 = 64 – 24 = 40.

(iv)

The document Quadratic Equations Class 10 Notes Maths Chapter 4 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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