If b is a real number and if a
2= b, then a
± =
‾¯√
.
Lecture (3) : Solving Quadratic Equations
Square Root Property
b
–
11
5x
2+ 55 = 0
x
± =
‾‾‾√
5x
2=
–
55
x
2=
–
11
x = ± i
‾‾‾√
11
(
3x – 1
)
2
= –4
3x – 1 = ±
√‾-4
3x – 1 = ± 2i
3x = 1 ± 2i
3
2
1
i
x
The quadratic formula is used to solve any quadratic equation
.
The quadratic
formula
is
:
Standard form of a quadratic equation is
:
Solving Quadratic Equations
The Quadratic Formula
2
2
2
4
2
x
b
b
a
c
a
2
0
x
b
x
The quadratic formula is used to solve any quadratic equation
.
The quadratic
formula
is
:
Standard form of a quadratic equation is
:
Solving Quadratic Equations
The Quadratic Formula
2
4
2
x
b
b
a
c
a
2
0
x
b
x
a
c
2
4
8 0
x
x
a
1
b
4
c
8
2
3
x
5
x
6 0
2
2
x
x
0
a
2
b
1
c
0
2
10
Solving Quadratic Equations
The Quadratic Formula
4
4
2
4
2
x
b
b
a
c
a
2
0
x
b
x
a
c
2
3
2 0
x
x
a
1
b
3
c
2
23
3
4
1
2
1
2
x
3
9 8
2
x
3
1
2
x
3 1
2
x
3 1
2
x
3 1
2
x
4
2
x
2
x
2
2
x
1
x
3 1
2
Solving Quadratic Equations
The Quadratic Formula
2
4
2
x
b
b
a
c
a
2
0
x
b
x
a
c
2
2
x
x
5 0
a
2
b
1
c
5
24
2
2
1
1
2
5
x
Solving Quadratic Equations
The Quadratic Formula
6
6
2
4
2
x
b
b
a
c
a
4
4
2
x
x
0
4
4
x
2
x
4
2
4
4
4
1
1
2
x
8
64
1
1
x
8
63
1
x
8
63
1
i
x
8
7
*
9
1
i
x
8
7
3
1
i
x
x
i
Solving Quadratic Equations
The Quadratic Formula and the Discriminate
The
discriminate
is the radicand portion of the quadratic
formula (b
2– 4ac)
.
It is used to discriminate among the possible number and type
of solutions a quadratic equation will have
.
b
2– 4ac
Name and Type of Solution
Positive
Zero
Two real solutions
One real solutions
2
4
2
x
b
b
a
c
a
Solving Quadratic Equations
The Quadratic Formula and the Discriminate
b
2– 4ac
Name and Type of Solution
Positive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real
solutions
Positive
Two real solutions
8
8
2
4
2
x
b
b
a
c
a
3
2
4
1
2
8
9
2
3
2 0
x
x
a
1
b
3
c
2
1
2
Solving Quadratic Equations
The Quadratic Formula and the Discriminate
b
2– 4ac
Name and Type of Solution
Positive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real
solutions
Negative
Two complex, non-real solutions
2
4
2
x
b
b
a
c
a
1
2
4
4
4
64
1
a
b
c
0
4
4
x
2
x
4
1
4
10
worksheet
worksheet
Copyright © by Houghton Mifflin Company, Inc. All rights reserved
. 14
A linear equation in one variable is an equation which can be written in the form:
ax + b = c
for a, b, and c real numbers with a 0
.
Linear equations in one variable
:
( 1 ) 2x + 3 = 11
( 2 ) 2 ( x 1
= ) 8
Not
linear equations in one variable
:
( 4 )
2x + 3y =
11
Two variables
can be rewritten 2x + (2) = 8
.
x
is squared
.
Variable in the denominator
( x 1
)
2 = 8
can be rewritten
x
+
5
=
7
.
14
7
5
3
2
)
3
(
x
x
3
1
7
5
3
2
A solution of a linear equation in one variable is a real number which, when substituted for the variable in the equation, makes the equation true.
Example: Is 3 a solution of 2x + 3 = 11?
2x + 3 = 11
2 ( 3 + ) 3 = 11
6 + 3 = 11
Original equation
Substitute 3 for
x
.
False equation
3
is not a solution of 2x + 3 = 11
.
Example: Is 4 a solution of 2x + 3 = 11?
2x + 3 = 11
2 ( 4 + ) 3 = 11
8 + 3 = 11
Original equation
Substitute 4 for
x
.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved
. 16
Addition Property of Equations
If
a
=
b
, then
a
+
c
=
b
+
c
and
a
c
=
b
c
.
Use these properties to solve linear equations.
Example: Solve x 5 = 12.
x 5 = 12
x 5 + 5 = 12 + 5
x = 17
17 5 = 12
Original equation
The solution is preserved when 5 is
added to both sides of the equation.
Check the answer.
17 is the solution.
That is, the same number can be added to or subtracted from each side of an equation without changing the solution of the equation.
Multiplication Property of Equations
If a = b and c 0, then ac = bc and .
That is, an equation can be multiplied or divided by the same nonzero real number without changing the solution of the equation.
Example: Solve 2x + 7 = 19. 2x + 7 = 19 2x + 7 7 = 19 7
2x = 12
x = 6
Original equation
The solution is preserved when 7 is
subtracted from both sides.
Simplify both sides.
6 is the solution.
The solution is preserved when each side
is multiplied by .
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
18
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Use the addition and subtraction properties to get all variable terms
on the left-hand side and all constant terms on the right-hand side.
3. Simplify both sides of the equation.
4. Divide both sides of the equation by the coefficient of the variable.
Example: Solve x + 1 = 3(x 5).x + 1 = 3(x 5)
x + 1 = 3x 15
x = 3x 16 2x = 16
x = 8 The solution is 8.
Check the solution:
Original equation
Simplify right-hand side.
Subtract 1 from both sides.
Subtract 3
x
from both sides.
Divide both sides by
2.
True
(8) + 1 = 3((8) 5) 9 = 3(3)
Example: Solve 3(x + 5) + 4 = 1 – 2(x + 6).
3(x + 5) + 4 = 1 – 2(x + 6)
3x + 15 + 4 = 1 – 2x – 12
3x + 19 = –2x – 11
3x = –2x – 30
5x = –30
x = 6 The solution is 6.
Original equation
Simplify.
Simplify.
Subtract 19.
Add 2
x.
Divide by 5.
Check.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
20
Equations with fractions can be simplified by multiplying both sides by a common denominator.
The lowest common denominator
of all fractions in the equation is 6.
3x + 4 = 2x + 8 3x = 2x + 4
x = 4
Multiply by 6.
Simplify.
Subtract 4.
Subtract 2
x.
Check.
True
Example: Solve .
4 4 6 6
20
)
4
(
3
1
3
2
2
1
x
x
4)
)
((
3
1
3
2
)
(
2
1
)
8
(
3
1
3
2
2
3
8
3
8
(
4
)
worksheet
ALGEBRAIC METHODS OF SOLVING
SIMULTANEOUS LINEAR EQUATIONS
The most commonly used algebraic methods of solving simultaneous
linear equations in two variables are:
1- Method of elimination by substitution
ELIMINATION BY SUBSTITUTION
STEPS :
Obtain the two equations. Let the equations be
a
1x + b
1y + c
1= 0 --- (i)
a
2x + b
2y + c
2= 0 --- (ii)
Choose either of the two equations, say (i) and find the
value of one variable , say ‘y’ in terms of x
Substitute the value of y, obtained in the previous step in
equation (ii) to get an equation in x
Solve the equation obtained in the
previous step to get the value of x.
Substitute the value of x and get the value
of y.
Let us take an example
SUBSTITUTION METHOD
x + 2y = -1
x = -2y -1 --- (iii)
Substituting the value of x in equation
(ii), we get
2x – 3y = 12
2 ( -2y – 1) – 3y = 12
- 4y – 2 – 3y = 12
- 7y = 14 , y = -2 ,
SUBSTITUTION
Putting the value of y in eq (iii), we get
x = - 2y -1
x = - 2 x (-2) – 1
= 4 – 1
= 3
Hence the solution of the equation is
ELIMINATION METHOD
•
In this method, we eliminate one of the two
variables to obtain an equation in one variable
which can easily be solved. Putting the value of
this variable in any of the given equations, the
value of the other variable can be obtained.
•
For example: we want to solve,
3x + 2y = 11
2x + 3y = 4
Let 3x + 2y = 11 --- (i)
2x + 3y = 4 ---(ii)
Multiply 3 in equation (i) and 2 in
equation (ii) and subtracting eq. iv
from iii, we get
9x + 6y = 33 --- (iii)
4x + 6y = 8 --- (iv)
5x = 25
=> x = 5
•
putting the value of y in equation
(ii) we get,
2x + 3y = 4
2 x 5 + 3y = 4
10 + 3y = 4
3y = 4 – 10
3y = - 6
y = - 2
worksheet
worksheet
Answer .No
u = 4 , z = 3 1
u = 2 , y = 5 2
c= 3 , u = 6 3
u = 2 , v = 6 4
a = 5 , x = 4 5
a = 6 , v = 1 6
b = 5 , v = 6 7
a = 4 , u = 1 8
