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Home > Quantitative aptitude questions and answers with explanation > Numbers > Important Concepts and Formulas - Algebra
Important Concepts and
Formulas - Numbers
Important Concepts and
Formulas - Algebra
Important Concepts and
Formulas - Sequence and
Series
Important Concepts and
Formulas - Logarithms
Important Concepts and
Formulas - Trigonometry
Important Concepts and
Formulas - Complex
Numbers
Solved Examples - Set 1
Solved Examples - Set 2Solved Examples - Set 3
Important Concepts and Formulas - Algebra
I. Basic Algebraic Formulas
1. (Distributive Law)
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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a ( b + c ) = a b
+ a c
( a + b = + 2 a b
+
)
2
a
2
b
2
( a − b = − 2 a b
+
)
2
a
2
b
2
( a + b + (
a − b = 2 ( + )
)
2
)
2
a
2
b
2
( a + b = + 3
b + 3
a + = + 3 a b
( a + b ) +
)
3
a
3
a
2
b
2
b
3
a
3
b
3
( a − b = − 3
b + 3
a − = − 3 a b
( a − b ) −
)
3
a
3
a
2
b
2
b
3
a
3
b
3
− = ( a − b
) ( a + b )
a
2
b
2
+ = ( a + b
) ( − a b + )
a
3
b
3
a
2
b
2
− = ( a − b
) ( + a b + )
a
3
b
3
a
2
b
2
− = ( a − b
) ( + b
+ + . . . + )
a
b
a
− 1
a
− 2
a
− 3
b
2
b
− 1
( a + b + c = + + + 2 ( a b
+ b c
+ c a
)
)
2
a
2
b
2
c
2
( a + b ) (
a + c ) = + (
b + c ) a + b c
a
2
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1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
1.
2.
3.
4.
5.
6.
7.
8.
= a . a . a . . . ( n t i m e s )
a
. . . . = a
a
a
a
+ + . . . +
. = a
a
a
+
=
a
a
a
−
= = ( ) a
a
( ) a
= ( a b
)
a
b
=
a
b
a
b
= a
−
1
a
= a
1
a
−
= 1 w h e r e a
∈ R , a ≠ 0 a
0
= a
/
a
− −
√
i f = w h e r e a
≠ 0 a n d a
≠ ± 1 , t h e n = a
a
i f = w h e r e
≠ 0 , t h e n a
= ± b a
b
(
+ a = + 2
a
+ )
2
2
a
2
( − a = − 2 a
+ )
2
2
a
2
( + a ) (
+ b ) = + (
a + b ) + a b
2
− = ( − a
) ( + a )
2
a
2
( + 1 ) i s c o m p l e t e l y d i v i s i b l e b y ( x + 1 ) w h e n n i s o d d
( + ) i s c o m p l e t e l y d i v i s i b l e b y ( x + a ) w h e n n i s o d d
a
( − ) i s c o m p l e t e l y d i v i s i b l e b y ( − a
) f o r e v e r y n a t u r a l n u m b e r n
a
( − ) i s c o m p l e t e l y d i v i s i b l e b y ( + a
) w h e n n i s e v e n
a
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Binomial Theorem
If a and b are any real numbers and n is a positive integer,
Binomial Coefficient also occurs in many other mathematical areas than algebra, especially in combinatorics wher
represents the number of combinations of n distinct things taking r at a time and is denoted by nCr or C(n,r).
The properties of binomial coefficients have led to extending its meaning beyond the basic case where n and r are nonn
integers with r ≤ n; such expressions are also called binomial coefficients.
Binomial Expansions
... and so on
Binomial Series
One of Newton's achievement was to extend the Binomial Theorem to the case where
n can be any real number.
If n is any real number and -1 < x < 1, then
II. Surds and Important Properties
i. Surds
Let be a rational number and be a positive integer such that is irrational.
Then is called a surd of order n.
( a + b )
= + b + + + ⋯ + + ⋯a
a
− 1
( − 1 )
2 !
a
− 2
b
2
( − 1 ) (
− 2 )
3 !
a
− 3
b
3
( − 1 ) (
− 2 ) ⋯ (
− + 1 )
!
a
−
b
= + b + + + ⋯ +
+ ⋯ +
0
a
1
a
− 1
2
a
− 2
b
2
3
a
− 3
b
3
a
−
b
b
= ∑
= 0
a
−
b
w h e r e
i s k n o w n a s B i n o m i a l C o e f f i c i e n t a n d i s d e f i n e d b y
= = w h e r e r = 1 , 2 , . . . , n a n d = 1
!
( ! ) (
− ) !
( − 1 ) (
− 2 ) ⋯ (
− + 1 )
!
0
( a + b = 1
)
0
( a + b = a + b )
1
( a + b = + 2 a b
+ )
2
a
2
b
2
( a + b = + 3
b + 3
a + )
3
a
3
a
2
b
2
b
3
( a + b = + 4
b + 6 + 4
a + )
4
a
4
a
3
a
2
b
2
b
3
b
4
( 1 + )
= 1 + + + + ⋯
(
− 1 )
2 !
2
(
− 1 ) (
− 2 )
3 !
3
= +
+
+
+ ⋯
0
0
1
1
2
2
3
3
= ∑
= 0
∞
w h e r e
= w h e r e r ≥ 1 a n d
= 1
( − 1 ) (
− 2 ) ⋯ (
− + 1 )
!
0
a a
√
a
√
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Examples :
is a surd of order 2
is a surd of order 3
is a surd of order 2
Please note that numbers like , etc are not surds because they are not
irrational numbers
Every surd is an irrational number. But every irrational number is not a surd. (eg
: , etc are not surds though they are irrational numbers.)
ii. Quadratic Surds
A surd of order 2 is called a quadratic surd
Examples :
iii. Rules of Surds
1.
2.
3.
4.
5. if
1. The conjugate surd of
2. if and are quadratic surds and if ,then and
i.e., rational part on the lef t side = rational part on the right side.
and irrational part on the left side = irrational part on the right side.
3. if and are quadratic surds and if ,
then and
III. Quadratic Equations and How to Solve Quadratic Equations
A. Quadratic Equations
A quadratic equation is a second degree univariate polynomial equation.
A quadratic equation can be written as (general form or standard form of quadratic
equation)
where x is a variable, a, b and c are constants and
Example :
(Please note that if a=0, equation becomes a linear equation)
The solutions of a quadratic equation are called its roots.
B. How to Solve Quadratic Equations
There are many methods to solve Quadratic equations. Quadratic equations can be
solved by factoring, completing the square, graphing, Newton's method, and using the
quadratic formula. We can go through some of the popular methods for solving
quadratic equations here.
3 √
7 √
3
4 3 √
9 √ 2 7
− −
√
3
π
, , ( 3 + ) , e t c . 2
√
3 √ 5
√
× =
− −
√
√
− − −
√
=
− −
√
√
− − −
√
= a . a . a ⋯ n t i m e s
− − − − − − − − − −
√
− − − − − − − − − − − − −
√
− − − − − − − − − − − − − − − − −
√
a
1 −
1
2
= a a
. a
. a ⋯ ∞
− − − − − −
√
− − − − − − − − −
√
− − − − − − − − − − − − −
√
= , t h e n
( − 1 ) = a a
+ a
+ a ⋯ ∞
− − − − − −
√
− − − − − − − − − − −
√
− − − − − − − − − − − − − − − −
√
+ = ± ( − ) a
√
b
√
a
√
b
√
b
√
√
a + = c +
b
√
√
a =
c b
=
b
√
√
a + =
b
√
√
a = 0
b =
a + b
+ c = 0
2
a ≠ 0
+ 5 + 6 = 0
2
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1. How to Solve Quadratic Equations By Factoring
Step 1: Write the quadratic equation in standard form.
i.e. , bring all terms on the left side of the equal sign and 0 on the right side of
the equal sign.
Step 2: Factor the left side of the equation
Step 3: Equate each factor to 0 and solve the equations
Factoring is one of the fastest ways for solving quadratic equations. The concept
will be clear from the following examples.
Example1 : Solve the Quadratic Equation
Step 1 : Write the quadratic equation in standard form.
Step 2 : Factor the left side of the equation.
We know that . We will use the same
concept for factoring.
Assume
We need to find out a and b such that a + b = 3 and ab = -10
a = +5 and b = -2 satisfies the above condition.
Hence
Step 3 :Equate each factor to 0 and solve the equations
(x + 5)(x - 2) = 0
=> (x + 5) = 0 or (x - 2) = 0
=> x = -5 or 2
Hence, the solutions of the quadratic equation are x = -5
and x = 2.
(In other words, x = -5 and x = 2 are the roots of the quadratic equation
Example2 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Here we need to find out a and b such that a + b = -7 and ab = +10
a = -5 and b = -2 satisfies the above condition.
Hence
Hence,
Step 3 :Equate each factor to 0 and solve the equations
(x - 5)(x - 2) = 0
=> (x - 5) = 0 or (x - 2) = 0
=> x = 5 or 2
Hence, the solutions of the quadratic equation are x = 5 and x
+ 4
− 1 1 =
− 1
2
+ 4 − 1 1 = − 1
2
⇒ + 4
− 1 1 −
+ 1 = 0
2
⇒ + 3
− 1 0 = 0
2
+ ( a + b ) +
a b = ( + a
) ( + b )
2
+ 3
− 1 0 = + ( a + b ) +
a b
2
2
+ 3
− 1 0 = (
+ 5 ) (
− 2 )
2
+ 3
− 1 0 = 0
2
⇒ (
+ 5 ) (
− 2 ) = 0
+ 4
− 1 1 =
− 1
2
+ 4
− 1 1 =
− 1
2
− 7
+ 1 0 = 0
2
− 7
+ 1 0 = (
− 5 ) (
− 2 )
2
+ 3
− 1 0 = 0
2
⇒ (
− 5 ) (
− 2 ) = 0
− 7
+ 1 0 = 0
2
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= 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
)
Example3 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Here we need to follow a slightly different approach for factoring because here
the coefficient of x2 ≠ 1 (whereas In example 1 and example 2, the coefficient of
x2 was 1)
Hence, to factor , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
2. We got product as 24x2 and have middle term as -14x. From this , it is clear
that we can take -12x and -2x such that their sum is -14x and product is 24x 2
Hence, =
Hence,
Step 3 :Equate each factor to 0 and solve the equations
(x - 5)(x - 2) = 0
=> (x - 4) = 0 or (3x - 2) = 0
=> x = or .
Hence, the solutions of the quadratic equation are x = and
x =
(In other words, x = and x = are the roots of the quadratic equation
Example4 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
coefficient of x2 ≠ 1
Hence, to factor , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
2. We got product as 72x2 and have middle term as -17x. From this , it is clear
that we can take -8x and -9x such that their sum is -17x and product is 72x 2
Hence,
Hence,
Step 3 :Equate each factor to 0 and solve the equations
(3x - 4)(2x - 3) = 0
=> (3x - 4) = 0 or (2x - 3) = 0
=> x = or .
Hence, the solutions of the quadratic equation are x =
and x = .
− 7
+ 1 0 = 0
2
3 − 1 4 + 8 = 0
2
3 − 1 4
+ 8
2
3 × 8 = 2 4
2
2
3 − 1 4
+ 8 = 0
2
3 − 1 2
− 2
+ 8 = 3 (
− 4 ) − 2 (
− 4 ) = (
− 4 ) ( 3
− 2 )
2
3 − 1 4
+ 8 = 0
2
⇒ (
− 4 ) ( 3
− 2 ) = 0
4
2
3
3 − 1 4
+ 8 = 0
2
4
2
3
4
2
3
3 − 1 4
+ 8 = 0
2
6 − 1 7
+ 1 2 = 0
2
6 − 1 7
+ 1 2
2
6 × 1 2 = 7 2
2
2
6 − 1 7
+ 1 2 = 6 − 8
− 9
+ 1 2 = 2
( 3
− 4 ) − 3 ( 3
− 4 ) = ( 3
− 4 ) ( 2
− 3 )
2
2
6 − 1 7
+ 1 2 = 0
2
⇒ ( 3
− 4 ) ( 2
− 3 ) = 0
4
3
3
2
6 − 1 7
+ 1 2 = 0
2
4
3
3
2
4 3
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(In other words, x = and x = are the roots of the quadratic equation
(Special Factorization Examples)
Example5 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Here, is in the form where a = x and b = 3
We know that We will use the same concept for
factoring this.
Hence,
Step 3 :Equate each factor to 0 and solve the equations
(x - 3)(x + 3) = 0
=> x = 3 or -3
Hence, the solutions of the quadratic equation are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
Example6 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Hence,
Step 3 : Equate each factor to 0 and solve the equations
(5x - 4)(5x + 4) = 0
=>5x = 4 or -4
.
Hence, the solutions of the quadratic equation are x = and x =
(In other words, x = and x = are the roots of the quadratic equation
Example7 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Here, is in the form where a = x and b = 3
We know that We will use the same concept for
factoring this.
2
4
3
3
2
6 − 1 7
+ 1 2 = 0
2
− 9 = 0
2
− 9
2
− a
2
b
2
− = ( a − b
) ( a + b ) a
2
b
2
− 9 = − = (
− 3 ) (
+ 3 )
2
2
3
2
− 9 = 0
2
⇒ (
− 3 ) (
+ 3 ) = 0
− 9 = 0
2
− 9 = 0
2
2 5 − 1 6 = 0
2
2 5 − 1 6 = ( 5
− = ( 5
− 4 ) ( 5
+ 4 )
2
)
2
4
2
2 5 − 1 6 = 0
2
⇒ ( 5
− 4 ) ( 5
+ 4 ) = 0
= o r
4
5
− 4
5
2 5 − 1 6 = 0
2
4
5
− 4
5
4
5
− 4
5
2 5 − 1 6 = 0
2
+ 6
+ 9 = 0
2
+ 6
+ 9
2
+ 2 a b + a
2
b
2
+ 2 a b + = ( a + b a
2
b
2
)
2
+ 6
+ 9 = + 2 ×
× 3 + = (
+ 3
2
2
3
2
)
2
+ 6
+ 9 = 0
2
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Hence,
Step 3 :
=> x + 3 = 0
=> x = -3
Hence, the solution of the quadratic equation is x = -3
(In other words, x = -3 is the root of the quadratic equation
Example8 : Solve the Quadratic Equation
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation.
Here, is in the form where a = x and b = 3
We know that We will use the same concept for
factoring this.
Hence,
Step 3 :
=> x - 3 = 0
=> x = 3
Hence, the solution of the quadratic equation is x = 3
(In other words, x = 3 is the root of the quadratic equation
2. How to Solve Quadratic Equations By Quadratic Formula
Quadratic Formula
Consider a quadratic equation where
Its solutions can be given by
Example1 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x = 2
and x = -5
(In other words, x = 2 and x = -5 are the roots of the quadratic equation
+ 6
+ 9 = 0
2
⇒ (
+ 3 = 0 )
2
( + 3 = 0
)
2
+ 6
+ 9 = 0
2
+ 6
+ 9 = 0
2
− 6
+ 9 = 0
2
− 6
+ 9
2
− 2 a b + a
2
b
2
− 2 a b + = ( a − b a
2
b
2
)
2
− 6
+ 9 = − 2 ×
× 3 + = (
− 3
2
2
3
2
)
2
+ 6
+ 9 = 0
2
⇒ (
− 3 = 0 )
2
( − 3 = 0
)
2
− 6 + 9 = 0
2
− 6
+ 9 = 0
2
a + b
+ c = 0
2
a ≠ 0
=
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
+ 4
− 1 1 =
− 1
2
+ 4
− 1 1 =
− 1
2
⇒ + 4
− 1 1 −
+ 1 = 0
2
⇒ + 3
− 1 0 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
− 3 ± − 4 × 1 × ( − 1 0 ) 3
2
− − − − − − − − − − − − − − −
√
2 × 1
− 3 ± 9 + 4 0
− − − − −
√
2
=
− 3 ± 7
2
= o r
4
2
− 1 0
2
= 2 o r − 5
+ 4 − 1 1 = − 1
2
+ 4
− 1 1 =
− 1
2
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Example2 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x = 5 and x
= 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
Example3 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x = and
x =
(In other words, x = and x = are the roots of the quadratic equation
Example4 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x =
and x = .
(In other words, x = and x = are the roots of the quadratic equation
Example5 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x =(-3 + i)
+ 4
− 1 1 =
− 1
− 7
+ 1 0 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
7 ± ( − 7 − 4 × 1 × 1 0
)
2
− − − − − − − − − − − − − − −
√
2 × 1
7 ± 4 9 − 4 0
− − − − − −
√
2
= =
7 ± 9 √
2
7 ± 3
2
= o r
1 0
2
4
2
= 5 o r 2
− 7
+ 1 0 = 0
2
− 7
+ 1 0 = 0 )
2
3 − 1 4
+ 8 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
1 4 ± ( − 1 4 − 4 × 3 × 8 )
2
− − − − − − − − − − − − − − −
√
2 × 3
1 4 ± 1 9 6 − 9 6
− − − − − − −
√
6
= =
1 4 ± 1 0 0
− − −
√
6
1 4 ± 1 0
6
= o r
2 4
6
4
6
= 4 o r
2
3
3 − 1 4
+ 8 = 0
2
4
2
3
4
2
3
3 − 1 4
+ 8 = 0
2
6 − 1 7
+ 1 2 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
1 7 ± ( − 1 7 − 4 × 6 × 1 2 )
2
− − − − − − − − − − − − − − − −
√
2 × 6
1 7 ± 2 8 9 − 2 8 8
− − − − − − − −
√
1 2
= =
1 7 ± 1
√
1 2
1 7 ± 1
1 2
= o r
1 8
1 2
1 6
1 2
= o r
3
2
4
3
6 − 1 7
+ 1 2 = 0
2
4
3
3
2
4
3
3
2
6 − 1 7
+ 1 2 = 0
2
+ 6 + 1 0 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
− 6 ± ( 6 − 4 × 1 × 1 0 )
2
− − − − − − − − − − − − − −
√
2 × 1
− 6 ± 3 6 − 4 0
− − − − − −
√
2
= =
− 6 ± − 4
− − −
√
2
− 6 ± 2
2
= − 3 ±
= ( − 3 +
) o r ( − 3 − )
+ 6
+ 1 0 = 0
2
-
8/18/2019 Important Concepts and Formulas - Algebra
10/11
11/15/2014 Important Concepts and Formulas - Algebra
http://www.careerbless.com/aptitude/qa/algebra_imp.php 10/12
and x = (-3 - i)
(In other words, x =(-3 + i) and x = (-3 - i) are the roots of the quadratic
equation
Example5 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation is
(In other words, is the roots of the quadratic equation
Example6 : Solve the Quadratic Equation
Hence, the solutions of the quadratic equation are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
C. How to find out the number of solutions(roots) of a quadratic
equation quickly?
To find out the number of solutions (roots) of a quadratic equation quickly, we need to find out the discriminant, D. The
discriminant, D =
If the discriminant is positive, there will be two real solutions for the quadratic
equation.
If the discriminant is 0, there will be one real solution (it repeats itself) for the
quadratic equation.
If the discriminant is negative, there will be two complex solutions for the
quadratic equation.
Example1 :
Find out the number of roots for
Discriminant, D =
D > 0. Hence will have two real roots.
Let's verify if this is correct by solving the quadratic equation.
Yes, two real roots we got.
Example2 :
Find out the number of roots for
Discriminant, D =
D = 0. Hence will have one real root (it repeats itself)
Let's verify if this is correct by solving the quadratic equation.
Yes, real root (it repeats itself).
+ 6
+ 1 0 = 0
2
+ 6 + 9 = 0
2
= = =
− b ± − 4
a c
b
2
− − − − − − −
√
2 a
− 6 ±
( − 6 − 4 × 1 × 9 )
2
− − − − − − − − − − − − − −
√
2 × 1
− 6 ±
( 3 6 − 3 6
− − − − − − −
√
2 × 1
= =
− 6 ± 0 √
2
− 6
2
= − 3
+ 6
+ 9 = 0
2
= − 3
= − 3 + 6
+ 9 = 0
2
− 9 = 0
2
= = =
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
0 ± ( 0 − 4 × 1 × ( − 9 ) )
2
− − − − − − − − − − − − − − − −
√
2 × 1
± 3 6
− −
√
2
= ±
6
2
= 3 o r − 3
− 9 = 0
2
− 9 = 0
2
a + b
+ c = 0
2
– 4 a c b
2
− 7
+ 1 2 = 0
2
– 4 a c = ( − ) − ( 4 × 1 × 1 2 ) = 4 9 − 4 8 = 1 . b
2
7
2
− 7
+ 1 2 = 0
2
− 7
+ 1 2 = 0 ⇒ (
− 3 ) (
− 4 ) = 0 ⇒
= 3 o r 4 .
2
− 4
+ 4 = 0
2
– 4 a c = ( − ) − ( 4 × 1 × 4 ) = 1 6 − 1 6 = 0 . b
2
4
2
− 4
+ 4 = 0
2
− 4
+ 4 = 0 ⇒ (
− 2 = 0 ⇒
= 2
2
)
2
-
8/18/2019 Important Concepts and Formulas - Algebra
11/11
11/15/2014 Important Concepts and Formulas - Algebra
Example3 :
Find out the number of roots for
Discriminant, D =
D < 0. Hence will have two complex roots
Let's verify if this is correct by solving the quadratic equation.
Yes, we got two complex roots, +i and -i
D. Sum and Products of the roots of a quadratic equation
Let and are the roots of a quadratic equation
Sum of the roots of the quadratic equation,
Product of the roots of the quadratic equation,
Example 1: Find the sum and product of the roots of the quadratic equation
sum of the roots
product of the roots
Example 2: Find the sum and product of the roots of the quadratic equation
sum of the roots
product of the roots
Comments(6) Newest
+ 1 = 0
2
– 4 a c = ( ) − ( 4 × 1 × 1 ) = 0 − 4 = − 4 b
2
0
2
+ 1 = 0
2
+ 1 = 0
2
= = = = = = ±
− b ± − 4 a c
b
2
− − − − − − −
√
2 a
0 ± − 4 × 1 × 1 0
2
− − − − − − − − − − − −
√
2 × 1
0 − 4
− − − − −
√
2
− 4
− − −
√
2
± 2
2
1
2
a + b
+ c = 0 w h e r e
a ≠ 0
2
+ =
1
2
− b
a
× =
1
2
c
a
− 7
+ 1 0 = 0
2
= = = 7
− b
a
7
1
= = = 1 0
c
a
1 0
1
2 + 3
+ 1 = 0
2
= =
− b
a
− 3
2
= =
c
a
1
2
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SRK 30 Sep 2014 12:12 PM
PLEASE HELP ME IN FINDING HCF AND LCM OF POLYNOMIALS
Reshma 12 Jun 2014 11:11 AM
very helpful site for students...:)
catty 25 May 2014 5:09 PM
so helpful guys! Thanks :)
Parthiban 03 Mar 2014 6:06 PM
This is help to me pls can you send me all the formulas
vitthal 13 Feb 2014 1:26 PM
give me problems
Usama 28 Sep 2013 12:18 PM
Can u mail me the formulae?
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