Quadratic equation

Quadratic Equations
Shivangi Tidke

ACKNOWLEDGEMENT
I would like to express my special thanks of gratitude to my teacher
Mr. Jal Engineer Sir as well as our principal Mr. N K Mishra Sir who
gave me the golden opportunity to do this wonderful project on the
topic Quadratic Equations , which also helped me in doing a lot of
Research and I came to know about so many new things I am really
thankful to them.
Secondly I would also like to thank my parents and friends who helped
me a lot in finalizing this project within the limited time frame.

INDEX
Sr no. Topics to be covered
1. Introduction
2. Quadratic Equation
3. Solving Quadratic equations by the square root property
4. Solving Quadratic equations by the factorization method
5. Solving Quadratic equations by completing the square method
6. Solving Quadratic equations by the quadratic formula
7. Summary

 Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x2– px + q = 0.
 Solving of quadratic equations, in general form, is often credited to ancient Indian
mathematicians. In fact, Brahmagupta (A.D.598–665) gave an explicit formula to solve a
quadratic equation of the form ax2+ bx = c.
 Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic
formula, as quoted by Bhaskara II) for solving a quadratic equation by the method
of completing the square.

Quadratic Equations
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. For example, 2x2 + x – 300 = 0 is a quadratic
equation. Similarly, 2x2– 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are
also quadratic equations. 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. Quadratic equations arise in several situations in the world around us
and in different fields of mathematics.

Solving Quadratic Equations by the
Square Root Property

If b is a real number and a2= b, then
Square Root Property
ba 
Example
Solve x2 + 4 = 0
x2 = 4
There is no real solution because the square root of 4 is not a real number.

Solve (y – 3) 2 = 4
y = 3  2
y = 1 or 5
243 y
Examples
Solve (x + 2) 2 = 25
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
5252 x
Solve (3x – 17)2 = 28
72173 x
3
7217 
x
7228 3x – 17 =

Solving Quadratic Equations by the
factorization method

In general, a real number α is called a root of
the quadratic equation ax2 + bx + c = 0, a ≠ 0
if aα2 + 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.

Solve x2 + 5x + 6 = 0.
This equation is already in the form "(quadratic) equals (zero)", this isn't yet factored. The quadratic must first be
factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and
solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6),
because you can add lots of stuff that totals zero.
So the first thing I have to do is to find factors through middle term split :
x2 + 5x + 6 = 0
x2 + 3x + 2x + 6 = 0
x (x + 3) +2 (x + 3) = 0
(x + 2) (x + 3) = 0
x + 2 = 0 or x + 3 = 0
x = –2 or x = – 3
The solution to x2 + 5x + 6 = 0 is x = –3, –2

Checking x = –3 and x = –2 in x2 + 5x + 6 = 0:
[–3]2 + 5[–3] + 6 = 0
9 – 15 + 6 = 0
9 + 6 – 15 = 0
15 – 15 = 0
0 = 0
[–2]2 + 5[–2] + 6 = 0
4 – 10 + 6 = 0
4 + 6 – 10 = 0
10 – 10 = 0
0 = 0

Solving Quadratic Equations by
completing the square

In all four of the previous examples, the constant in the square
on the right side, is half the coefficient of the x term on the
left.
Also, the constant on the left is the square of the constant on
the right.
So, to find the constant term of a perfect square
trinomial, we need to take the square of half the
coefficient of the x term in the trinomial (as long as
the coefficient of the x2 term is 1, as in our previous
examples).

What constant term should be added to the following expressions to create a perfect square
trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
add
4
49
2
7
2







• We now look at a method for solving
quadratics that involves a technique called
completing the square.
• It involves creating a trinomial that is a
perfect square, setting the factored trinomial
equal to a constant, then using the square root
property from the previous section.

Solving a Quadratic Equation by Completing a Square
1) If the coefficient of x2 is not 1, divide both sides of the equation by the
coefficient.
2) Isolate all variable terms on one side of the equation.
3) Complete the square (half the coefficient of the x term squared, added to both
sides of the equation).
4) Factor the resulting trinomial.
5) Use the square root property.

Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y = 3 ± 1
y = 4 or 2
y + 3 = ± = ± 11

Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
2
29
4
29
2
1
y 2
291
2
29
2
1 
y
(y + ½) 2 = 4
29

Solving Quadratic Equations by
quadratic formula

Another technique for solving quadratic equations is to use the quadratic formula.
The formula is derived from completing the square of a general quadratic equation.
A quadratic equation written in standard form, ax2 + bx + c = 0, has
the solutions.
a
acbb
x
2
42



Consider the quadratic equation ax2 + bx + c = 0
(a ≠ 0). Dividing throughout by a, we get

Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1



)11(2
)1)(11(4)9(9 2
n 

22
44819


22
1259
22
559 




)1(2
)20)(1(4)8(8 2
x 

2
80648


2
1448


2
128
20 4
or , 10 or 2
2 2


x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
8
1
2
5
Solve x2 + x – = 0 by the quadratic formula.

The expression under the radical sign in the formula (b2 – 4ac) is
called the discriminant.
The discriminant will take on a value that is positive, 0, or
negative.
The value of the discriminant indicates two distinct real
solutions, one real solution, or no real solutions, respectively.
The Discriminant

Use the discriminant to determine the number and type of solutions
for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.

Summary
1. A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠
0.
2. A real number α is said to be a root of the quadratic equation ax2 + bx + c = 0, if aα2 + bα + c = 0. The
zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are
the same.
3. If we can factorize ax2 + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic
equation ax2 + bx + c = 0 can be found by equating each factor to zero.
4. A quadratic equation can also be solved by the method of completing the square.
5. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 are given by
provided b2– 4ac ≥ 0.
6. A quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2– 4ac > 0,
(ii) two equal roots (i.e., coincident roots), if b2 – 4ac = 0, and
(iii) no real roots, if b2 – 4ac < 0.

BIBLIOGRAPHY
• mathworld.wolfram.com/QuadraticEquation
• https://www.mathsisfun.com/algebra/quadratic-equation
• https://en.wikipedia.org/wiki/Quadratic_equation
• mathworld.wolfram.com › ... › Interactive Entries › Interactive Demonstrations
• https://www.khanacademy.org/math/algebra/quadratics
• www.purplemath.com/modules/solvquad4.htm
• www.themathpage.com/alg/quadratic-equations.htm

Quadratic equation

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