This document discusses quadratic equations. It defines a quadratic equation as having degree 2 in the standard form ax2 + bx + c = 0. It provides examples of quadratic equations and explains that the roots are the values that satisfy the equation. Methods for solving quadratic equations are outlined, including factorization, completing the square, and the quadratic formula. The quadratic formula is defined as x = (-b ± √(b2 - 4ac))/2a. Discriminant is also discussed, which is denoted by D and equals b2 - 4ac, and how it relates to the number of real roots.

1. Guided By: Bharti Mam
Submitted By: RAGSYA GROUP
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2.  Rishabh Bahadur
 Shubham Vishwakarma
 Rahul Parpani
 Yogin Chetwani
 Yash Singh
 Sumit Manjhi
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3.  An Equation can be said as a
Quadratic Equation if its degree is 2 .
 The Standard form of a Quadratic
equation is
where a,b,c are integers and a≠0
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4.  6x² + 11x – 35 = 0
 2x² – 4x – 2 = 0
 2x² – 64 = 0
 x² – 16 = 0
 2x² + 8x = 0
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5.  The roots of a quadratic equation are the values of
the variable that satisfy the equation.
They are also known as the "solutions" or
"zeros" of the quadratic equation.
For example, the roots of the quadratic equation
x2 - 7x + 10 = 0 are x = 2 and x = 5 because they
satisfy the equation.
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7.  Step 1: Consider the quadratic equation ax2 + bx + c
= 0
 Step 2: Now, find two numbers such that their
product is equal to ac and sum equals to b.
(number 1)(number 2) = ac
(number 1) + (number 2) = b
 Step 3: Now, split the middle term using these two
numbers, ax2 + (number 1)x + (number 2)x + c = 0
 Step 4: Take the common factors out and simplify. 7

8. x2 + 8x + 12 = 0
⇒ x2 + 6x + 2x + 12 = 0
Now, club the terms in pairs as:
(x2 + 6x) + (2x + 12) = 0
⇒ x(x + 6) + 2(x + 6) = 0
Taking the common factor (x + 6) out, we have
(x + 2) (x + 6) = 0
Thus, (x + 2) and (x + 6) are the factors of x2 + 8x + 12 = 0
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9.  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 the sides
 Step 6: Solve for variable x and find the roots. 9

10. Example :- Find the roots of the
quadratic equation x2 + 4x – 5 = 0
Given quadratic equation is:
x2 + 4x – 5 = 0
Comparing the equation with the standard form,
b = 4, c = -5
(x + b/2)2 = -(c – b2/4)
So, [x + (4/2)]2 = -[-5 – (42/4)]
(x + 2)2 = 5 + 4 ⇒ (x + 2)2 = 9
⇒ (x + 2) = ±√9 ⇒ (x + 2) = ± 3
⇒ x + 2 = 3, x + 2 = -3 ⇒ x = 1 , -5
Therefore, the roots of the given equation are 1 and -5. 10

11.  The quadratic formula is a rule that says that ,
in any equation of the form , ax2 + bx + c = 0,
the solution X-values of the equation are
given by:
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12. 1. Arrange your equation into a standard form
2. Pull out the numerical parts of each of these
terms
3. , which are the ‘a’ , ’b’ , ’c’ of the formula .
4. Plug these numbers into the formula
5. Simplify and get your answer
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14. Question: Factorize x2 + 4x – 21 =0
using quadratic formula.
x2 + 4x – 21 = 0
Here, a = 1, b = 4, c = -21
b2 – 4ac = (4)2 – 4(1)(-21) = 16 + 84 = 100
Substituting these values in the quadratic formula, we get;
x = [-4 ± √ (4)2 – 4(1)(-21)]/ 2(1)
= (-4 ± 10)/2
x = (-4 + 10)/2 , x = (-4 – 10)/2
x = 6/2 , x = -14/2
x = 3 , x = -7
Therefore, the factors of quadratic equation are (x – 3) and (x + 7).
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15. Discriminant is denoted by , ‘D’
Its formula is = b²-4ac
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D<0; No Real roots exist

18. i. A quadratic equation in the variable x is of the form of
ax2 + bx + c = 0 where a , b , c are real no.s and a ≠ 0
ii. Roots of the quadratic equations are also called
zeroes of the polynomial who satisfy the equation
iii. There are 3 methods to solve a quadratic equation
iv. Factorization , completing the square , quadratic
formula .
v. 4. Quadratic formulaa
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