Chapter 4 : Quadratic equation | CBSE board | 10th class board | Ncert solution

Chapter 4 : Quadratic equation

What is quadratic equation

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.
Example 2x² + x – 300 = 0 , 2x² – 3x + 1 = 0, 4x – 3x² + 2 = 0 and 1 – x² + 300 = 0
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax² + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

How to check whether given equation is quadratic equation or not.

Point 1 : After simplify the equation, highest power of variable should be 2.
Point 2: After simplify the equation , equation can be represented in the form ax²+bx+c = 0
Example : a) (x – 2)² + 1 = 2x – 3
(x – 2)² + 1 = 2x – 3 can be rewritten as
x² – 4x + 5 = 2x – 3 i.e.,
x² – 6x + 8 = 0
It is of the form ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation.
b) (x + 2)³ = x³ – 4
(x + 2)³ = x³ + 6x² + 12x + 8
(x + 2)³ = x³ – 4 can be rewritten as
x³ + 6x² + 12x + 8 = x³ – 4
6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation

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Solution of Quadratic Equation:

a real number α is called a root of the quadratic equation ax² + bx + c = 0, a ≠ 0 if a α² + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic equation ax² + bx + c = 0 are the same.
Method:
(i) factorization
(ii) Quadratic formula
a) factorization
1. 2x² – 5x + 3 = 0
Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x² = (2x²) × 3].
So, 2x² – 5x + 3 = 0
2x² – 2x – 3x + 3 = 0
2x(x – 1) –3(x – 1) = 0
(2x – 3)(x – 1) = 0
Now, 2x² – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x² – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives x = 3/2
and x – 1 = 0 gives x = 1.
x = 3/2 and x = 1 are the solutions of the equation.
2. 6x² – x – 2 = 0.
Solution : We have
6x² – x – 2 = 0
6x² + 3x – 4x – 2 = 0
3x(2x + 1) – 2 (2x + 1) = 0
(3x – 2)(2x + 1) = 0
The roots of 6x² – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0
Therefore, 3x – 2 = 0 or 2x + 1 = 0,
i.e.x = 2/3
and x = -1/2
Therefore, the roots of 6x² – x – 2 = 0 are -1/2 and 2/3
b) Quadratic formula
The equation ax² + bx + c = 0 are given by

So , there are two roots

Example:
1. 2x² – 3x - 5 = 0
Use the quadratic formula.
a = 2, b = -3, c = -5

Nature of roots:

Using the discriminant, nature of roots can be determined.
Suppose a quadratic equation is given in the form of ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0. It has
(i) two distinct real roots, if b² – 4ac > 0
(ii) two equal roots (i.e., coincident roots), if b² – 4ac = 0
(iii) no real roots, if b² – 4ac < 0.
Question :
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
First find the value of discriminant ( b² – 4ac)
a = 2 , b = -3 , c = 5
now , put the value of a , b and c in discriminant formulla
= b² – 4ac
= (-3)² - 4(2)(5)
= 9 - 40
= -31
the value of discriminant is negative , so no real roots is possible

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