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7 questions on Quadratic Equations by Factoring :
1) Using the factoring method, solve for the roots of each equation. Place
equation in standard form before factoring. Check your solutions and show
the check.
3x2 = 75
Solution:

Divide both sides by 3.
 x 2 - 25 = 0
Subtract 25 on both sides.
x2 - 52
= 0
(x+5)(x-5) = 0

x 2 = 25
since 5 2 = 25
since
a 2 - b 2 = (a + b) (a - b)
x = -5 or x = 5 are the solutions.
We then check its solutions by substitution method .
replacing x = 5  3x2 = 3 (5)2
replacing x = -5 
= 3 * 25 = 75 which is the right side.
3x2 = 3(-5)2
= 3 * 25 = 75.
2) Using the factoring method, solve for the roots of each equation. Place
equation in standard form before factoring. Check your solutions and show
the check.
4x2 + 3x = 4x
Solution:
Given our equation as : 4x2 + 3x = 4x
Subtract 4x on both sides.

4x2 + 3x – 4x = 4x – 4x

4x2 – x
Taking x term common outside ,
 x = 0
 x =0
or
or
x ( 4x -1 ) = 0
(4x-1) = 0

4x = 1
check their values using substitution method .
x = ¼
are its solutions.
replacing x = 0
replacing x = ¼
therefore x = ¼


4(0) – 3(0) = 4(0)
0 = 0 therefore



4 (¼)2 + 3( ¼ ) = 4( ¼ )
4 ( 1/16 ) + ¾ = 1
¼+¾=1

x=0
is a solution.
1=1
is also a solution.
3) Using the factoring method, solve for the roots of each equation. Place
equation in standard form before factoring. Check your solutions and show
the check.
9x2 + 4x + 1 = 10x
Solution:
9x2 + 4x + 1 = 10 x
Subtract 10x on both sides.

9x2 + 4x -10x + 1 = 10x – 10x
9x2 – 6x + 1 = 0
writing -6x as -3x – 3x

Taking the common terms out :
9x2 -3x- 3x + 1 = 0
3x from the first two terms and -1 from the next two,
 3x (x-1) – 1 (3x-1) = 0
Then take (x-1) common from both the terms

x –1=0
or
Now resolving for x from them , we get :

x = 1
or
 (x - 1) ( 3x - 1) = 0
(3x-1) = 0
x=1
or
3x = 1
x = 1/3
we then check for its solution by substitution method
replacing x = 1

9x2 + 4x + 1 = 10x
9(1) + 4(1) + 1 = 10 (1)
9 + 4 + 1 = 10
which is not true.
Therefore x = 1 is not a solution.
Again replacing x = 1/3



9x2 + 4x + 1 = 10x
9 (1/3)2 + 4 (1/3) + 1 = 10 (1/3)
1 + 4/3 + 1 = 10/3
3 4  3
= 10/3
 10 /3 = 10 /3
3

That makes x = 1/3 as it’s solution.
4) Using the factoring method, solve for the roots of each equation. Place
equation in standard form before factoring. Check your solutions and show
the check.
(x2/4) – 1 – 2x = – 1
Solution:


Add 1 on both sides.
x2/4 – 1 – 2x + 1 = -1 + 1
x2/4 – 2x = 0
Taking common terms ‘ x ‘ outside , we get :
Equating them to zero



x=0
x=0
x=0
or
or
or
x ( x/4 – 2 ) = 0
x/4 – 2 = 0
x/4 = 2
x = 8
We then check for its solution, using substitution method .
replacing x = 0 
(0) – 1 – 2(0) = -1

-1 = -1
Therefore x = 0 is a solution.
replacing x = 8 
therefore x = 8
(64/4) -1 – 2(8) = -1

16 -1 -16 = -1
is also a solution.
5) Using the factoring method, solve for the roots of each equation. Place
equation in standard form before factoring. Check your solutions and show
the check.
x2 + 5x = – 2
3
Solution:
Multiply both sides of equal to sign by 3 , we get

Add 6 on both sides,
then by factoring method , we get
 x 2 + 5x = -6
x2 + 5x + 6 = 0
:
( x + 2) ( x + 3) = 0
solving them for x , we get their values as :
x = -2
or
x = -3
We then check its solution by substitution method ,
replacing x = -2
therefore x = -2
4 10
= - 6/3 = -2 , which is the right side.
3
is a solution.

9 15
= -6/3 = -2
3
that makes x = -3 also to be a solution of it .
then replacing x = -3

6) Suppose the number of students in a mathematics class is x. The teacher
insists that each student participate in group work each class. The number of
possible groups is: G = x2 – 3x + 2
2
The class has 17 students. How many possible groups are there?
Solution :
Given x = 17 , so replacing it in our equation
G = x2 – 3x + 2
2
=
(17)2 – 3(17) + 2
2
=
120
That makes for 17 students , number of possible groups = 120
7) The technology and communication office of a local company has set up a
new telephone system so that each employee has a separate telephone and
extension number. They are studying the possible number of telephone calls
that can be made from people in the office to other people in the office. They
have discovered that the total number of possible telephone calls T is
described by the equation T = 0.5(x2 – x), where x is the number of people in
the office. If 90 people are presently employed at the office, how many
possible telephone calls can be made between these 90 people?
Solution :
 x = 90
Given , there 90 people employed
So replacing x = 90 in our equation , we get
T = 0.5 ( 902 – 90 ) = 4005
:
That makes , the number of telephone calls possible as 4005 .
7 questions on Quadratic Equations by the square root property :
8) Solve using the square root property; express any complex numbers using i
notation: x2 + 25 = 0
since

we have our expression as x2 + 25 = 0
Solution :
i2 = -1
, we get
Simplifying it further

x2 = 25 i 2

x 2 = ( 5 i )2
Now by taking square root to both the sides ,
x = + 5i
and
x2 = - 25
we get
x = -5i
as its solutions .
9) Solve using the square root property; express any complex numbers using i
notation: (7x + 2) 2 = 3
Solution :
we have our expression as :
(7x + 2) 2 = 3
Now taking square root for both the sides ,we get
that splits as :
7x + 2
= 3
or
7x + 2 =
7x + 2
 3
= - 3
resolving them for x , we get their values as :
x
= ( 3 - 2 ) / 7
or
x
= - ( 3 + 2 ) / 7
10) Solve the equation by completing the square; completely simplify your
answer; express any complex numbers using i notation: x2 – 6x = 13
Solution :
we have our expression as x2 – 6x = 13
( x2 – 6x ) + 9 = 13 + 9
Now add 9 to both the sides of equal to sign :

using

a2 – 2ab + b2 = ( a – b ) 2
( x – 3 )2
x2 – 6x + 9
with a = x
x - 3
and b = 3 , we get :
= 22
now taking square roots to both the sides gives us :

= 22
= 22
or
x - 3

= - 22
now taking 3 to the other side , we get their solutions as
x = 3 + 22
( x – 3 ) =  22
:
x = 3 - 22
or
11) Solve the equation by completing the square; completely simplify your
answer; express any complex numbers using i notation: 3y2 – 4y = 4
Solutions :
we have our expression as 3y2 – 4y = 4
( 3y2 – 4y ) + 4/3 = 13 + 4/3
Now add 4/3 to both the sides of equal to sign :

using

a2 – 2ab + b2 = ( a – b ) 2
(3y – 2/3 ) 2
= 43/3
with a = 3y and b = 2/3 , we get :
= 43/3
now taking square roots to both the sides gives us :

3y2 – 4y + 4/3
3y – 2/3 = (43/3)
or

( 3y – 2/3 ) =  (43/3)
3y – 2/3 = - (43/3)
now multiplying by 3 to the other side , we get their expressions as
:
y - 2
= 43
Further resolving for y ,
or
we get
= - 43
y - 2
y = 2 + 43
:
y = 2 - 43
or
12) Solve the equation by completing the square; completely simplify your
answer; express any complex numbers using i notation:
2x2 + 12x – 7 = – 2
Solutions :
2x2 + 12x – 7 = – 2
we have our expression as
now dividing it by 2 , we get :

now adding 9 to both the sides
then regroup the terms
x2 + 6x – 7/2 = – 1
( x2 + 6x – 7/2 ) + 9

( x2 + 6x + 9 ) - 7/2

( x + 3 ) 2 - 7/2
bringing 7/2 to the other side


x + 3
solving for x ,

x
=
(23/2)
or
= 8
= 8
( x + 3 )2
now taking square root for both the sides , we get
= 8
:
x + 3
= 8 + 7/2
= 23/2
( x + 3 ) =  (23/2)
= - (23/2)
we get their solutions as :
= (23/2) - 3
x = - ( (23/2) + 3 )
or
13) Use either the square root property or completing the square. Express any
complex numbers using i notation: [(x/3) + 4]2 = 27
Solutions :
we have our expression as
[(x/3) + 4]2 = 27
Now taking square root to both the sides

(x/3) + 4
then subtracting 4 from both the sides

x/3
=  33
- 4
then multiplying by 3 to both the sides

x
=  93
- 12
=
 33
that is :
x
= 93
- 12
or
x
= - ( 93
+ 12 )
14) The time a basketball player spends in the air when shooting a basket is
called "the hang time". The vertical leap L measured in feet is related to the
hang time t measured in seconds by the equation L = 4t2. A typical athlete
has a vertical leap of 1 1/2 to 2 feet; the best male jumpers attain heights of 3
1/2 to 4 feet. What is the hang time of an athlete who has a vertical leap of
3.9 feet?
Solutions :
Given the vertical leap , L = 3.9 ft
then dividing by 4 , makes

L = 4 t2
replacing it in our relation :
:
t2
take square root to both the sides :
= 3.9/4
t =
4 t2
= 3.9
= 0.975
0.987 seconds
as the hang time .
6 questions on solving a quadratic equations by using the
quadratic formula :
Note :
For a quadratic equation of kind :
its solution as
x 
a x2 + b x + c = 0 , we get
b 
b 2  4a c
2a
15) Solve by the quadratic formula; simplify your answer; use i notation to
express any nonreal complex numbers. Clearly label your values for a, b,
and c.
9x2 – 12x – 1 = 0
Solutions :
From the above note , we have
a = 9 , b = - 12 ,
c = -1
x 
So from the quadratic formula :
 x
=
 x
=
12 
144  36
18
2 
5
12 2  4 9  1
12 
2 9
12 
=
180
18

12  6
5
18
as its solution.
3
16) Solve by the quadratic formula; simplify your answer; use i notation to
express any nonreal complex numbers. Clearly label your values for a, b,
and c.
4x2 – 10 = 1
Solutions :

Subtract 1 from both the sides , to have the right side expression as zero
4x2 – 11 = 0
that makes
=
 x
=
 4
11
=
8

11

x 
So from the quadratic formula :
 x
a = 4 , b = 0 ,

c = - 11
 4 4  11
2 4
11
2
as its solution.
2
17) Solve by the quadratic formula; simplify your answer; use i notation to
express any nonreal complex numbers. Clearly label your values for a, b,
and c.
x(x + 3) = (– 4x – 11)/4
Solutions :
that makes :
multiplying by 4 to both the sides
4x2 + 12 x
Simplifying further gives :
 4x ( x + 3 ) = - 4x - 11
= - 4x - 11
4x2 + 16 x + 11
so from the above note , we have
= 0
a = 4 , b = 16 ,
c = 11
So from the quadratic formula :
 x
=
 x
=
4 
x 
:
16 2  4 4 11
 16 
2 4
16  11
2
4 
5
as its solution
2
18) Solve by the quadratic formula; simplify your answer; use i notation to
express any nonreal complex numbers. Clearly label your values for a, b,
and c.
1_ + 3
12
x
= _2_
x+1
Solutions :

multiplying by 12 x ( x + 1 ) , we get our expression as
x ( x + 1 ) + 36 ( x + 1 ) =
Simplifying it further , would give
x2 + x + 36x + 36 = 24 x
:

that makes
a = 1 , b = 13 ,
So from the quadratic formula :
 x =
 x =
 13 
2
-9
or
x
x2 + 13x
+ 36
= 0
c = 36
:
169  144
24 x
=
= -4
13 2  4 1 36
 13 
x 
 13 
25
2
2 1
=
 13  5
2
as its solution
19) Solve by the quadratic formula; simplify your answer; use i notation to
express any nonreal complex numbers. Clearly label your values for a, b,
and c.
x2 – 6x + 25 = 0
Solutions :
From the above note , we have
So from the quadratic formula :
 x =
6 
 64
2
 x = 3 + 4i
and
=
a = 1
x 
:
6 
64 i 2
2
x
= 3 - 4i
6 
=
, b= -6
, c = 25
36  100
2 1
6  8i
2
as its solutions.
20) Use the discriminant to find what type of solutions (two rational, two irrational,
one rational, or two nonreal complex) the equation has. Do not solve the
equation.
4x2 + 9 = 12x
Solutions :
we have our expression as
4x2 + 9 = 12x
subtract 12x from both the sides , to have right side zero
that makes :
Discriminant
a = 4
of it
b = - 12
=

4x2 - 12 x + 9 = 0
c = 9
b 2  4a c
=
144  4 49
=
0
when discriminate equals zero , we get exactly one real number solution to exist.
Hope this helps !!