This document discusses quadratic equations and methods for solving them. It defines quadratic equations and lists objectives like methods for finding solutions, the discriminant, nature of roots, and examples. It then explains three methods for finding solutions: factorization, completing the square, and the quadratic formula. It provides examples of using each method and defines key terms like discriminant and the nature of roots based on the discriminant's value.

1. QUADRATIC EQUATION
CLASS 10TH

2. OBJECTIVES
 QUADRATIC EQUATION DEFINITION
 METHODS TO FIND THE SOLUTION OF
QUADRATIC EQUATIONS
 DISCRIMINANT
 NATURE OF ROOTS
 EXAMAPLES

3. QUADRATIC EQUATION
A polynomial p(x) = ax2 + bx + c of degree 2 is called a
quadratic polynomial, then p(x) = 0 is known as a quadratic
equation.
e.g. 2x2 – 3x + 2 = 0, x2 + 5x + 6 = 0 are quadratic equations.

4. METHODS TO FIND THE SOLUTION OF
QUADRATIC EQUATIONS
Three methods to find the solution of a quadratic equation:
1.Factorization method
2. Method of completing the square
3. Quadratic formula method


5. FACTORISATION METHOD
Steps to find the solution of a given quadratic equation by factorization:
Firstly, write the given quadratic equation in standard form ax2 + bx + c = 0.
Find two numbers α and β such that the sum of α and β is equal to b and the product
of α and β is equal to ac.
Write the middle term bx as αx + β x and factorize it by splitting the middle term and
let factors are (x + p) and (x + q) i.e. ax2 + bx + c = 0  (x + p)(x + q) = 0
Now equate the reach factor to zero and find the values of x.
These values of x are the required roots/solutions of the given quadratic
equation.

6. Find the roots of the quadratic equation 6x2 – x – 2 = 0
Solution : We have 6x2 – x – 2 = 6x2 + 3x – 4x – 2
= x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
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 or x = -1/2
Therefore, the roots of 6x2 – x – 2 = 0 are 2/3 and –1/2
We verify the roots, by checking that 2/3 and -1/2 satisfy 6x2 – x – 2 = 0

7. METHOD OF COMPLETING THE SQUARE
Steps for Completing the Square Method:
Suppose ax2 + bx + c = 0 is the given quadratic equation. Then follow
the given steps to solve it by completing the square method.
 Step 1: Write the equation in the form, such that c is on the right
side.
 Step 2: If a is not equal to 1, divide the complete equation by a
such that the coefficient of x2 will be 1.
 Step 3: Now add the square of half of the coefficient of term-x,
(b/2a)2, on both sides.
 Step 4: Factorize the left side of the equation as the square of the
binomial term.
 Step 5: Take the square root on both sides

8. The above steps can be implemented as shown below.
Consider the quadratic equation ax2 + bx + c = 0 (a ≠ 0).
Dividing throughout by a, we get
x2 + (b/a)x + (c/a) = 0
This can also be written by adding and subtracting (b/2a)2 as,
[x + (b/2a)]2
– (b/2a)2
+ (c/a) = 0
[x + (b/2a)]2
– [(b2 – 4ac)/4a2] = 0
[x + (b/2a)]2
= [(b2 – 4ac)/4a2]
If b2 – 4ac ≥ 0, then by taking the square root, we get
x + (b/2a) = ± √(b2 – 4ac)/ 2a
Further simplification of this will give you the quadratic
formula.

9. Example: Solve 4x2 + x = 3 by completing the square
method.
Solution: Given,
4x2 + x = 3
Divide the whole equation by 4.
x2 + x/4 = ¾
Coefficient of x is ¼
Half of ¼ = ⅛
Square of ⅛ = (⅛)
2
Add (⅛)2 on both sides of the equation.
x2 + x/4 + (⅛)2 = ¾ – (⅛)2
x2 + x/4 + 1/64 = ¾ + 1/64
(x + ⅛)
2
= (48+1)/64 = 49/64
Taking square root on both sides, we get;
x+1/8 = √(49/64) = ±7/8
x = ⅞ – ⅛ = 6/8 = ⅜
x = – ⅞ – ⅛ = -8/8 = -1

10. QUADRATIC FORMULA METHOD
If b2 – 4ac ≥ 0, then the roots of the quadratic equation
ax2 + bx + c = 0 are given by
x = [-b ± √(b2 –
4ac)]/ 2a
This formula for finding the roots of a quadratic equation
is known as the quadratic formula.

11. Solve x2 – 5x + 6 = 0.
Solution: Given,
x2 – 5x + 6 = 0
a = 1, b = -5, c = 6
By the quadratic formula, we know;
x = [-b ± √(b2 – 4ac)]/ 2a
b2 – 4ac = (-5)2 – 4 × 1 × 6 = 25 – 24 = 1 > 0
Thus, the roots are real.
Hence,
x = [-b ± √(b2 – 4ac)]/ 2a
= [-(-5) ± √1]/ 2(1)
= [5 ± 1]/ 2
i.e. x = (5 + 1)/2 and x = (5 – 1)/2
x = 6/2, x = 4/2
x = 3, 2
Therefore, the solution of x2 – 5x + 6 = 0 is 3 or 2.

12. DISCRIMINANT
 The quadratic equation ax2 + bx + c = 0 has real roots or not,
b2 – 4ac is called the discriminant of this quadratic equation.

13. Determine the discriminant value for the given quadratic
equation 2x2+8x+8.
Solution:
Given: The quadratic equation is 2x2+8x+8
Here, the coefficients are:
a = 2, b = 8 and c = 8
The formula to find the discriminant value is D = b2 –
4ac
Now, substitute the values in the formula
Discriminant, D = 82 – 4(2)(8)
D = 64 – 4 (16)
D = 64 – 64
D = 0
The discriminant value is 0

14. NATURE OF ROOTS
A quadratic equation ax2 + bx + c = 0 has
(i) Two distinct real roots, if b2 – 4ac > 0,
(ii) Two equal real roots, if b2 – 4ac = 0,
(iii) No real roots, if b2 – 4ac < 0.

15. Find the discriminant of the quadratic equation 2x2 – 4x + 3 =
0, and hence find the nature of its roots.
Solution: 2x2 – 4x + 3 = 0
The given equation is of the form ax2 + bx + c = 0,
where a = 2, b = – 4
and c = 3
Therefore, the discriminant b2 – 4ac
= (– 4)2 – (4 × 2 × 3)
= 16 – 24 = – 8 < 0
So, the given equation has no real roots.

16. THE END