## Quadratic Expressions and Equations

Algebraic Expressions

Algebraic expressions is a mathematical phrase in which numbers and letters are connected by mathematical symbols.

Examples of algebraic expressions are

a + b, 2c + 1, q + 3q + r, 3a + 2b + 2c ' 5a

When algebraic expressions involve fractions they are referred to as algebraic fractions. Examples of these fractions are (where the denominator )

Algebraic expressions can be expanded or factorized.

**Algebraic Expressions**

Algebraic expressions is a mathematical phrase
in which numbers and letters are connected by mathematical symbols.

Examples of algebraic expressions are

a + b, 2c + 1, q + 3q + r, 3a + 2b + 2c ' 5a

When algebraic expressions involve fractions they are referred to as
algebraic fractions. Examples of these fractions are
(where
the denominator
)

Algebraic expressions can be expanded or factorized.

**Expansion of algebraic expressions**

Worked out example

Expand 3 (2x + 1)

**Solution **

3(2x + 1) = 3 x 2x + 3 x 1

= 6x + 3

**Note:** that in expansion the brackets are removed by
multiplying the value(s) outside the bracket by each of the values in
the brackets.

**Example 2**

Expand

2x (x + 2a + 6b)

*solution*

2x ( x + 2a + 6b) = 2x x 2 + 2x x 2a + 2x + 6b

2x2 + 4xa + 12xb

**Example 3**

Expand and simplify(a + 2) (3a ' 5)

Solution

= a x 3a = a x 5 + 2 x 3a ' 2 x 5

= 3a2 ' 5a + 6a ' 10

= 3a2 + a ' 10

**Factorization **

In algebraic expressions factorization is writing the expression as a
product of its factors.

We do this by introducing brackets and writing the common factor (s)
outside the brackets.

Worked example

**Example 1**

Factorize (6x + 4y)

*Solution*

The common factor is 2

Therefore 6x + 4y = 2(3x + 2y)

**Example 2**

Factorise

6x + 12y ' 24n ' 36m

**Solution **

The common factor is 6

Therefore 6x + 12y ' 24n ' 36m

= 6(x + 2y ' 4n -6m)

**Example 3**

Factorise

Mn ' 2m + 5n ' 10

**Solution **

Mn ' 2m _ 5n ' 10

Group the terms with common factors together as follows

(mn ' 2m) + (5n ' 10)

M(n - 2) + 5(n - 2) (Factor out the common factors outside the brackets)

(n ' 2) is a common factor

Therefore when factored out we get (n-2) (m+5)

(animate (n ' 2) by blicking in different color)

__Exercise__

Use factorization to solve for x in

X2 ' 9x- 36 = 0

__Solution__

__Step1__

x2+3x-12x ' 36 = 0

x (x+3) ' 12 (x+3) = 0

(x-12)(x+3) = 0

Either (x+3) = 0 just voice x = -3

Or (x-12) 0 then x =12

2. Solve for y in the quadratic equation

6x2 + 11x ' 20 =15

__Solution__

6x2 + 11x -35 = 0

6x2 + 21x ' 10x ' 35 = 0

3x(2x+7) ' 5 (2x+7) = 0

(3x-5) (2x+7) = 0

Either 3x-5 = 0 Voiced 3x = 5

and X= 5/3

Or 2x + 7 = 0 Voiced 2x = -7

and x= -7/2

3. The base of a right angled triangle is 7cm less than its height(h). If the area of the triangle is 99cm2.Find the dimensions of the triangle.

__Solution__

From the diagram A= ' x h x b

__h__

(h-7)

Therefore, ' x h x h-7= 99cm2

=h2 -7h = 198

=h2 -7h ' 198 = 0

=h2 +11h - 18(h+11) = 0

=(h ' 18) (h+11)= 0

**h **=** 18 or h **= -**11**

The height of the triangle = 18cm

Therefore, the base of the triangle = 18-7

=**11**

One thousand shillings is supposed to be shared equally among a number
of people .If the number of people is reduced by 10 each of the
remaining persons world receive fifty shillings more. Find how much each
person world receive if nobody withdraws from the group.

__Solution__

Let the number of people by x.

x people receive

x -10 people receive

=
50

Multiplying both sides by the L.C.M

1000x ' 1000(x-10) =

1000x ' 1000x + 10000 = 50x2 ' 500x

50x2 ' 500x ' 10,000 = 0

X2 ' 10x ' 200 =0

Solving the quadratic equation

X2 ' 20x + 10(x-20) = 0

X2 ' 20x +10x ' 200 = 0

X(x-20) + 10(x-20) = 0

(x-20) (x+10) = 0

x = 20 or x= -10

Since the number of people cannot be negative we take x = 20

Therefore, if nobody withdraws from the group each person would receive

=sh
50

**Objectives **

#### By the end of the topic you should be able to

:solve quadratic equations by factorization

form and solve quadratic equations**Introduction **

Some problems have more than one solution and
has to choose the most convenient answer depending on the situation.

Fig 1.

Animate movement form A to C

And from A through B and then to C

To move from A to C, one can decide to go direct or move from A to B and then to C.

**Lesson development **

**Introduction **

An algebraic expression that contain two terms is called a binomial
expression e.g.

(a + b), (2 ' c) (x + y) etc.

**Expansion of two binomial expressions with
similar variables gives a quadratic expressions**

An algebraic expression that contain two terms is called a binomial
expression e.g.

(a + b), (2 ' c) (x + y) etc.

**Example 1**

Expand (x + 1) (x + 2)

*Solution *

(x + 1) (x + 2)

x (x + 2) + 1 (x + 2) Splitting the first bracket i.e. (x + 1)

x2 + 2x + x + 2 (Removing brackets)

x2 + 3x + 2 (Collecting like terms)

x2 + 3x + 2 (This is a quadratic expression)

**Example 2**

Expand (20 + 3) (p ' 2)

*Solution*

(2p + 3) (p ' 2)

= 2p (p -2) + 3 (p-2)

= 2p2 ' 4p + 3p ' 6

= 2p2 ' p ' 6

**NB:** An expression of the form ax2 + bx + c where
the highest power of the variable is 2 is called a quadratic expression

a is the coefficient of x2

b is the coefficient of x

c is the constant term

The following quadratic identities are useful in expansion of binomial
expressions.

- (a + b) 2 = (a + b) (a + b)

= a (a + b) + b(a + b)

= a2 + ab + ab + b2

= a2+ 2ab + b2

- (a ' b) 2 = (a ' b) (a ' b)

= a (a ' b) ' b(a ' b)

= a2 ' ab ' ab + b2

= a2 ' 2ab + b2

- (a2 + b2) (a ' b)

= a (a ' b) + b (a ' b)

= a2 ' ab + ab ' b2

but -ab + ab = 0

Therefore (a + b) (a - b) = a2 ' b2 (This is called a difference of two
squares)

**Solving quadratic
equations using factorization**

When a quadratic expression is equated to zero we get a standard form of
the quadratic equation i.e. (ax2 + bx + c = 0 (Highlight the equation)

However the expression can be equated to any other value e.g. 2x2 + 5x +
3 = 4.

**Worked examples**

**Example 1**

Solved the following quadratic equations

4x2 + x = 0

*Solution *

When 4x2 + x is factorized we get x(4x + 1)

Therefore for 4x + x = 0

x (4x + 1) = 0 which means

Either

=x = 0

or 4x + 1 = 0

4x + 1 = 0

subtracting
1 form both sides

Therefore the solutions of x in 4x2 + x = 0 are x = 0 and x = - '

**Example 2**

x2 + 8x + 12 = 0

*Solution *

x2 + 6x + 2x + 12 = 0 (splitting the middle term into two number whose
and products is

1 x 12 = 12)

x (x + 6) + 2 (x + 6) = 0

(x + 2) (x + 6) = 0

Either x + 2 = 0 x = -2

Or x + 6 = 0 x = -6

The solutions of x in x2 + 8x + 12 = 0 are x = -2 and x = -6

**Example 3**

iii) 3y2 ' 19y ' 14 = 0

*Solution *

3y2 + 2y ' 21y ' 14 = 0

= y (3y + 2) -7 (3y + 2) = 0

(y - 7) (3y + 2) = 0

Either y ' 7 = 0

y = 7

Or 3y + 2 = 0 3y = - 2 (dividing both sides by 3)

y =

Therefore the solution of x in 3y2 ' 19y ' 14 = 0

Are y = 7 and y =

**Example 4**

? k2 + 5/4 k + 2 = 0

k2 + 10k + 16 = 0 (Multiplying through by the lcm of 8 and 4)

k2 + 8k + 2k + 16=0

k(k + 8) + 2(k + 8) = 0

(k + 2) (k + 8) = 0

Either k + 2 = 0 and wiced k = - 2

Or k + 8 = 0 and void k = -8

Therefore solutions of k in ? k2 + 5/4 k + 2 = 0

Are k = 2 and k = -8

**Example 5**

v) Solve for t in 8t2 + 14t + 1 = -2

*Solution*

8t2 + 14t + 3 = 0 (adding 2 on both sides to form a standard
quadratic equation)

8t2 + 12t + 2t + 3 = 0

4t(2t + 3) + 1 (2t + 3) = 0

(4t + 1) (2t + 3) = 0

Either 4t + 1 + 0

Or 2t + 3 = 0

The solutions of t is 8t2 + 14t + 1 = -2 are t = - ' and t = - 1'

**Example 6**

Solve for d in 1/9d2 ' 1 = 0

d2 ' 9 = 0 (multiplying all through by the LCM -9)

(d + 3) (d-3) = 0 (Difference of two squares)

Either d + 3 = 0 and voiced d = -3

Or d ' 3 = 0 and voiced d = +3

Therefore the difference for d in the equation 1/9d2-1 = 0

Are d = 3 and d = -3

**Note**

Generally from the examples above quadratic equations have two
solutions. The solutions of quadratic equations are also called the roots
of the equation.

**Forming and solving quadratic equation**

We can form quadratic equations from their roots as follows : -

Solution

Either x ' 2 = 0 or x ' 3 = 0

Therefore (x ' 2) (x ' 3) = 0 (finding the product of the two binomials)

x (x ' 3) ' 2(x-3)=0(Expanding)

x2-3x-2x+=0 ( collecting the like terms)

x2 =5x+6=0

Therefore, our quadratic equation is x2-5x+6=0

**Example 2**

Form the quadratic equation from the roots given. P =1/2 as p=2/3

__Solution__

Either(P - 1/2) (P - 2/3)=0

P (P - 2/3) - 1/2(P - 2/3)=0

P2 - 2/3p-1/2p+1/3 =0

P2 - 7/6p+1/3=0

**Example 3**

Find the quadratic equation whose roots are q= -10 and q = 1/3

__Solution__

Either q+10 = 0 or q-1/3 = 0

(q+10)
(q - 1/3) = 0

= q(q - 1/3) +10 (q - 1/3) = 0

= q2 - 1/3q +10q - 10/3 = 0

= 3q2 ' q + 30q ' 10 = 0 (Multiplying through by 3)

= 3q2 + 29q ' 10 = 0

The
quadratic equation from the roots is

3q2+29q-10 = 0

Quadratic equation can also be formed from real life situations.

This can be illustrated by the examples given below.

**Example 1**

A piece of barbed wire is used to fence a rectangular field whose area
is 300m2. Given that the length of the rectangular field is 10m more
than twice its width (i) find the width of the field.

(ii)The length of the barbed wire.

__Solution__

(2y+10)

y

- Let the value of the width be y m.
- Then the length will be (2y+10)

Therefore, the area of the rectangle is LxW=A

y x (2y+10)=300

2y2+10y=300

2y2+10y-300=0 (simplifying the equation dividing through by 2)

y2 + 5y ' 150 = 0

Factorizing

y2+5y-150=0

y2+15y-10y-150 = 0

y2+15y-10y-150 = 0

y(y+15) -10(y+15) = 0

y(y -10) (y +15) = 0

Either y -10=0 where y=10

Or

y + 15=0 where y= -15

We cannot have negative measurements and therefore the value of y= 10m

The
width of the field is 10m

The length of the field is 2(10)+10= 30m

The perimeter of the rectangular field is

= 2(L) + 2(W)

= 2(30) + 2(10)

= 60 + 20

= 80m

Therefore the length of the barbed wire is __80m__

__Example 2__

The distance from Nairobi to Eldoret is 360km. A bus and a Nissan left
Nairobi at 8.00am and travelled towards Eldoret. If the average speed of
the Nissan was 20km/h more than that of the bus and that it arrived at
Eldoret 3hours ealier than the Bus.

Find (i) The speed of the

(ii)The time the bus arrived at Eldoret.

__Solution__

From the information given, let the speed of the Nissan be km/h and
the speed of the Bus will be (S-20) km/h since the distance from Nairobi
to Eldoret=360km.

Then the time taken by the Nissan will be

The time taken by the Bus will be

Difference in their time of arrival = 3hrs

Therefore,

Multiplying by the L.C.M S(S ' 20) we get

360(S) - (360)(S-20) = 3(S)(S-20)

360S ' 360S+7200 =3(S2-20S)

360S-7200 = 3S2-60S

3S2-60S-7200 = 0 (forming a standard quadratic equation and simplify)

S2 ' 20S -2400 = 0

S(S-60) + 40(S-60) =0

(S+40)(S-60) =0

Either
S+40 = 0 where S= -40

Or S-60 = 0 where S=60

The speed cannot be negative therefore we take the positive value of S

The speed of the Nissan S = 60km/h

The speed of the bus will be 60km/h -20km/h

= 40km/h

Time taken by the bus will be

__360 __= __9hrs__

40

From 8.00 am we add 9hrs

8.00am + 9hrs = 17.00hrs

= __5.00p.m__

Quadratic Expressions and Equations

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