WEEK 5
Factorization of Quadratic Expressions.
Factorize the following:
- 6a2 + 15a + 9
- 6a2 – 19ax – 36x2
- (5x – 1) ( x – 3) – ( x – 5) ( x – 3)
- 35 – 2b – b2
- x2 – y2 + ( x + y ) 2
- 25a2 – 4 ( a – 2b ) 2.
Solutions.
6a2 + 15x + 9.
Since 3 is a common factor to all the terms, first take out 3 as the common factor:
6a2 + 15a + 9 = 3 (2a2+ 5a + 3)
= 3 (2a2 + 2a + 3a + 3)
= 3 (2a (a+1) + 3 (a+1)
= 3 ( (a+1) (2a+ 3)).
Hence.
6a2 + 15a + 9 = 3 (a+ 1) (2a + 3)
= 3 (a+1) ( 2a + 3)
- 6a2 – 19ax – 36x2
1st step: Find the product of the first and last terms.
6a2 x -36x2 = -216a2x2
2nd step: Find two terms such that their products is – 216a2x and their sum is -19ax ( the middle term).
Factors of -216a2x2 sum of factors .
a. +27ax and -8ax + 19ax
b. -27ax and +8ax – 19ax
c. +9ac and -24ax – 15ax
d. -9x and 24ax + 15ax.
Of these only b gives the required result.
3rd step: Replace -19ax in the given expression by -27ax and 8ax then factorize by grouping:
6a2 – 19ax – 36x2
= 6a2 – 27ax + 8ax – 36x2
= 3a (2a – 9x) + 4x (2a-9x)
=(2a – 9x ) ( 3a + 4x)
hence,
6a2 – 19ax – 36x2 = (2a -9x) (3a + 4x)
- (5x – 1 (x -3) – (x -5)(x – 3)
= (x – 3) ( 5x -1) – (x – 5)
= ( x -3) ( 5x – 1 – x + 5 )
= ( x – 3) ( 5x – x + 5 – 1)
= ( x – 3) ( 4x + 4)
= ( x – 3) + ( x + 1)
= ( x -3) 4(x+1)
= 4(x-3) (x+1)
4. 35 – 2b –b2
or – b2 – 2b + 35
= -b2 – 7b + 5b + 35
= -b (b+7) + 5 (b + 7)
= (b+7) ( -b + 5)
= (b+7) ( 5-b)
or (7+b) ( 5 – b)
5. x2 – y2 + (x + y)2
Since
X2 – y2 = (x)2 – (y)2
= ( x + y) ( x –y)
then x2 – y2 + (x + y)2 = (x + y)(x-y) + (x+ y)2
= (x +y)( x-y + (x + y)
= (x + y ( x –y + x + y)
= ( x + y ) ( x + x
= (x + y) (2x)
(x +y ) (2x) = 2x ( x +y)
EVALUATION.
Factorize the following
- m2 – 15mm – 54n2
- 8a2 – 18b2
- If 17x = 37 52 – 3562
Find the value of x
ASSIGNMENT
Factorize the following :
- a2 + 5ab – 36b2
- r2 – 25
- 10p2 – 41p – 45
- 12m2 – 4mn -5n2
- 12a2b2 + 11ab – 5.
Theory.
Factorize the following
- 8c2 – 14c – 9
- x2y2 – xy – 30.
If I is a root of the equation 5x2 + kx – 3 = 0
Find the other root.
3. Given that ½ and -3 are the roots of the equation 0 = ax2 + bx + C, find a, b,c, where a, b and c are the least possible integers.
Solutions
- Since 2 ½ is a root of the equation, then (x – 2 ½ ) is a factor, similarly, if -1 is a root of the equation then (x – (1-1) is a factor. i.e (x + 1) is a factor.
Thus, the required equation is : (x – 2 ½ ) ( x + 1) = 0
x (x + 1) – 2 ½ ( x + 1) = 0
x2 + x – 2 ½ x – 2 ½ = 0
x2 – 1 ½ x – 2 ½ = 0.
X2 – 3/2 x – 5/2 =0
2x2 – 3x – 5 =0
2. Since I is a root of the given equation
5x2 + kx – 3 = 0
then it must satisfy the given equation.
Substitute I for x,
5(1)2 + k(1) – 3 = 0
5 x 1 + k – 3 = 0
5 + k – 3 = 0
\5-3 + k = 0
2 + k = 0
k = 0-2
k = -2.
Hence the given equation becomes
5x2 – 2x – 3 = 0
by factorization method, this quadratic equations can be solved as follows :
5x2 – 2x – 3 = 0
5x2 – 5x + 3x – 3 = 0
5x (x -1) + 3 ( x -1) = 0
(x-1) ( 5x + 3) = 0
(x -1) = 0 or 5x + 3 = 0
i.e. x = 0+ 1 or 5x = 0-3
x = 1 or 5x = -3
x = 1 or x = –3/5.
Hence the other roots of the given quadratic equation is x = 3/5.
- Since the given roots of the equation 0 = ax2 + bx + C are ½ and -3, then the factors are (x – ½ ) and (x – c – 3)
i.e. ( x – ½ ) and ( x + 3)
Thus, the required equation is
(x – ½ ) ( x + 3) = 0
x(x+ 3) – ½ (x + 3) = 0
x2 + 3x – ½ x – 3/2 = 0
2x2 + 6x – x – 3 = 0
2x2 + 5x – 3 = 0
Comparing this with the given equation ax2 + bx + C = 0
Then
A= 2, b= 5, c = -3.
EVALUATION
I. 7 and -3 are the roots of the quadratic equation x2 +kx -21 = 0,what is the value of k?
2. -5 is a root of the
ASSIGNMENT
1.Find the quadratic equation whose roots are 2 and -3.
(a) (x2 + x -6 (b) x2 – x -6 (c ) x2 + 5x – 6 (d) x2 – 5x + 6 (e) x2 + 3x – 6.
2. (2x + 3) is a factor of 6x2 + – 12. The other factor is ………….
(a) (x + 6) (b) (2x – 3) (c) (3x + 4) (d) (3x – 4) (e) (4x – 9).
3. Find the roots of the equation x2 + 12x – 28 = 0 the
Examples
- The electrical resistance R ohms of a wire varies directly as the length cm and inversely as the square roots of the diameter d cm3.
- Express d in terms of l,R and the constant of variation k.
- Find the value of d, correct to 2 decimal places, when 1 = 15cm, R = 0.13ohms and k = 1.25 x 10-3.
- V varies jointly as the square of x and inversely as y, if V =18 when x = 3 and y =4, find V when x = 6 and y =9.
Solutions.
The greater of the two roots is …………………………………
(a) -14 (b) -2 (c ) 2 (d) 14
4. What is the product of the roots of the equation x2 -3x + 2 =0?
(a) -3 (b) -1 (c) 2 (d ) 3 (e)5
5. If x2 – 10x -24 =0, the sum of the roots of the equation is ………….
(a) – 10 (b) 10 (c ) -24 (d) -2
THEORY
- find the quadratic equation whose roots are (3 + √5).
- If 1 is root of the equation 5×2 + kx – 3 =0, find the other root.
Quadratic equation n2 – 8n – 65 =0, what is the other root of the equation?
Topic: Solution of Quadratic Equation by Factorization Method
Examples:
Solve the following quadratic equations by the factorization methods.
- m2 = 11m + 42
- 2x2 – x – 21 =0
- 2p2 + 11p = 30.
Solutions.
(1 ) m2 = 11m + 42
then m2 – 11m – 42 =0
1st step: Find the product of the first and last terms :
m2 x – 42 =-42m2
2nd step:find two terms such that their product is -42m2 and their sum is -11m ( the middle term).
Factors of -42m2 sum of factor s
(a) + 3m and – 14m – 11m
(b) – 6m and +7m + 11m
( c) -6m and + 7m +m
(d) +6m and -7m -m
Out of these, only (a) gives the required result.
3rd step: Replace -11 in the given expression by +3m and -14m. then factorise by grouping:
m2 -11m -42 = 0.
M2 + 3m – 14m – 42 =0
M(m+ 3) (m – 14) = 0
M + 3 = 0 or m – 14 =0
i.e m = 0 -3, or m = 0 + 14.
M = -3 or m = 14.
2. 2x2 –x -21 =0
1st Step: Product of first and last terms = 2x2 x -21
= – 42x2
2nd step: Find two terms such that their product is 42x2 and their sum is –x (the middle term).
Factors of-42x2 sum of factors
(a) + 2x and -21x -19x
(b) -2x and 21x +19x
( c)-3x and + 14x + 11x
(d) + 3x and -14x -11x
( e) +6x and -7x -x
(f) -6x and + 7x +x
out of these, only (e)gives the required result.
3rd step: Replace –x in the given expression by +6x and -7x. then factorize by grouping
2x2 – x -21 =0
2x2 +6x -7x – 21 = 0
x ( x + 3) -7 ( x + 3 ) = 0
( x + 3) ( x-7) =0
either x + 3 = 0 or 2x -7 =0
i.e x = 0-3 or 2x = 7
x = -3 or x =7/2
x = -3 or x = 3 ½ .
3. 2p2 + 11p = 30.
2p2 + 11p – 30 = 0
1st step: Product of the first and last terms
2p2 x -30 =- 60p2
2nd step: find two terms such that their products is -60p2 and their sum is +11p (the middle term )
Factors of +11 sum of factors
(a) + 6p and – 10p -4p
(b) -15p and +4p +4p
( c) -15p and +4p -11p
(d) +15p and -4p +11p.
Out of these, only (d) gives the more required solution. Replace + 11p by +15p and -4p in the given expression .
2p2 + 15p -4p -30 = 0
p ( 2p + 15) – 2 (2p + 15) = 0
(2p + 15) (p – 2) = 0
either 2p+ 15 = 0 or p -2 = 0
2p = 0 – 15 or p = 0 + 2
2p = -15 or p = 2
p = -15 or p = 2
p = –15/2 or p = 2
p = -7 ½ or p = 2.
EVALUATION.
Solve the following quadratic equations by factorization method.
- 4e2 – 20 e + 25 = 0
- 4a2 – 11a = 3
ASSIGNMENT.
1. If x2 – 10 x – 24 = 0, then x = 12 or ……………
(a) -3 (b) -2 (c) -1 (d) 1 (e) 2
2. if x2 + Kx + 16/9 is a perfect square, then K = ………………..
(a) 1/3 (b) 2/3 (c) 4/2 (d) 8/3 (e) 16/3
3. What is the sum of the roots of the equation
x2 -3x +2 =0?
(a) – 3 (b) -1 ( c) 2 (d )3 (e) 5
4. Find the roots of the equation
x2 + 12x – 28 = 0
The greater of the two roots is ………….
(a) -14 (b) -2 (c ) 2(d)7 (e ) 14.
5. What are the factors of 6×2 + x – 12?
A.(2x+ 3)( x + 6) B. (2x +3) ( 2x – 3) C. (2x +3)(3x + 4) D. (2x + 3) (3x -4)
E. (2x + 3 ) ( 4x -9).
THEORY.
Solve the following equations”
1. 4y2 – 28y + 49 =0
2. 2x2 + 11x + 5 = 0.
Completing the square method
Given a quadratic equation,
a +bx=0 (where a, b and c are constants)
a +bx =-c (subtract c from both sides)
+ (divide both sides by a)
The coefficient of x=
Divide by 2 =
Square and add to both sides i.e
+ = i.e.
(X + = or
X +
X +
X =
X = . Q.E.D.
4Y + 4 = 0
SOLUTION
4Y = 4 (Re-arrange the equation)
(Divide through by coefficient of )
4Y+ ( = . Half Y, square it and add it to both sides.
=
(Y = 0 Y2 =
Y = 0 +2 Y = +2 or 2
OR using method of difference of two squares
(Y2)(Y+2) = 0
Y 2 = 0 or Y + 2 = 0
Y = 2 or Y = 2.
Example – Solve the equation using completing the square method.
(1) X2 – 8x + 3 = 0
(2) 3x2 – 5x – 7 = 0
Solution
1. x2 – 8x + 3 = 0
Step 1: Take C to the other side x2 – 8x = -3
Step 2: Divide through by the coefficient of x2
x2 – 8x = -3
1 1 1
Step 3: Divide the coefficient of x by 2, then square and add it to both sides.
Co efficient of x = 8
: 8 = 4
2
42 = 16
=x2 – 8x + (-8)2 = -3 + (- 8)2
2 2
x2 – 8x + (-3)2 = -3 + (-4)2
(x – 4)2 = -3 + 16
(x – 4)2 = 13
x – 4 = + 13
x = +4 + 13
x = 4 + 3.61
x = 4 – 3.61 or 4 + 3.61
0.39 or 7.61
2. 3x2 – 5x – 7 = 0
3x2 – 5x = +7
3x2 – 5x = 7
3 3 3
x2 – 5x = 7
3 3
x2 – 5x + (-5)2 = 7 + (-5)2
3 6 3 6
(x – 5)2 = 7 + 25
6 3 36
(x – 5)2 = 84 + 25
36
(x -5)2 = +109
6 36
x = 5 + +109
6 6
x – 5 = +10.44
6 6
x – 5 = +10.44
6 6
x = 5 + 10.44 or 5 – 10.44
6 6
x = 15.44 or -5.44
6 6
x = 2.5.573 or – 0.906
= 2.57 or – 0.91
m = 7 + 2.24
2
m = 7 + 2.24 or 7 – 2.24
2 2
m = 9.24 or 5.24
2 2
m = 4.62 or 2.62
2. 3x2 + 7x -3 = 0
3x2 + 7x = 3
3x2 + 7x = 3
3 3 3
x2 + 7x = 1
3
x2 + 7x + (7)2 = 1 + (7)2
3 6 6
(x + 7)2 = 1 + 49
6 36
(x + 7)2 = 1 + 36 + 49
6 36
(x + 7)2 = 85
6 36
x + 7 = +85
6 36
x + 7 = +85
6 6
x + 7 = +9.22
6 6
x = -7 + 9.22 or -7 – 9.22
6 6
Examples
Solve the following equations by completing the square method.
- n2 – 12n + 1 = 0
- y2 +7y – 30 =0
- m2 -7m + 11 = 0
Solutions
1. n2 – 12n + 1 = 0
n2 – 12n = 0 – 1
n2 – 12n = -1.
Add to both sides the square of –12/2 ( -6).
n2 – 12n ( -6) 2 = -1 + (-6)2
(n – 6) 2 = -1 + (-6) 2
(n – 6) 2 =+ 35.
Take square root of both sides.
n-6 = ± √35.
i.e n = ± √35 + 6
n = + √35 + 6 or n = -√35 + 6
n = 6 + √35 or n = 6 – √35
n = 6 + 5.916 or n = 6 – 5.916
n = 11.916 or n = 0.084.
i. e . n = 11.92 or n = 0.08 to 2 decimal places.
2.y2 +7y – 30 = 0
y2 + 7y = 0 + 30
y2 + 7y = 30
Add to both side the square of 7/2
y2 +7y + ( 7/2)2 = 30 + (7/2) 2
i.e. ( y +7/2) 2 = 30+ 49
1 4.
( y + 7/2) 2 = 120 + 49
4
( y + 7/2)2 = 169
4.
Take square root of both sides :
:. Y + 7/2 = ± √169/4/
y + 7/2 = ± 13/2
i.e y = ± 13/2 –7/2
.e y = +13/2 – 7/2 or y = – 13- 7
2.
y = +6 or y = -20
2 2
y = +3 or y = -10.
3.m2 – 7m + 11 = 0
m2 – 7m = 0 – 11
m2 -7m = -11
Add to both sides the square of –7/2
M2 – 7m ( –7/2)2 = -11 + (-7/2)2
( m – 7/2)2 = -11 + 49
4
( m – 7/2)2 = -44 + 49
4.
( m – 7/2)2 = 5/4
Take square root of both sides:
m –7/2 = ± √5/4
m= ±√5/4 + 7/2
m=± √+ 7/2
i.e. m = + √5 + 7 or m = – √5 + 7
2 2 2 2.
m = + 2.236 +7 or m = – 2.236 + 7
2 2 2 2
m=+1.118 + 3.5 or m = -1.118 + 3. 5
m = + 4.618 or m = 2.418
i.e m = 4.62orm = 2.42.
to 2 decimal places
EVALUATION.
Solve the following method of completing the squares:
- x2 – 8x – 1 = 0
- p2 -10p + 15 = 0
WEEKEND ASSIGNMENT
1. The greater of the two roots of the equations (2x -5) (3x +10 ) = 0 is
(a) – 50 (b) -10 ( c) -3 (d) 2 ½ ( e) 5.
2. If one of the roots of the equation ( n-13)2 = 9 is 10, what is the other root?
(a) 4 (b) 16 ( c) 22 (d) 68 ( e ) 94.
3. Find the value of k such that x2 + 5x +k is a perfect square.
(a) 2 ½ (b) 4 ¼ ( c ) 6 ¼ ( d) 25 ( e) 100.
4. If x2 – 10x = 24, what is the values of x ?
(a) 12,-3) (b) ( -2, 12) ( c) -1, 12) (d) ( 1, 12) (e) (2,12).
5. What must be added to a2 + 3/5a to make it a perfect square?
(a) 3/10 (b) 6/5 ( c) 9/100 (d) 100/9 ( e) 36/5
THEORY
Solve the following equations, giving answers correct to 1 decimal place.
- x2 + 14x + 8 = 0
- 2p2 +9p – 30 =0
Quadratic Formula
Quadratic formula is derived by using the method of completing the square. Considering the general form of quadratic equation:
ax2 + bx + c = 0 X = The required quadratic formula.
Example – Solve x2 + 3x – 2 = 0 using formula method
a = 1 6 = 3 C = – 2
x = -b b2 – 4ac
2a
x = -3 32 – 4(i) (-2)
2(i)
x = -3 9 + 8
2
x = -3 17
2
x = – 3 4.12
2
x = -3 + 4.12 or -3 – 4.12
2 2
x = 0.56 or – 3.56
2. m2 – 7m + 11 = 0
Solve using formula method.
m2 – 7m + 11 = 0
a = 1 b = – 7 c = 11
m = – b b2 – 4ac
2a
m = – (-7) 72 – 4(i) (ii)
2(i)
m = 7 49 – 44
2
m = 7 5
2
m = 7 2.23
2
m = 7 + 2.23 or 7 – 2.23
2 2
m = 4.62 or 2.62
EVALUATION
1. 3x2 + 7x – 3 = 0 solve using formula method
2. Using completing the square and formula method solve 3x2 – 12x + 10 = 0
Sum & Product of Roots of a Quadratic Equation
The expression for sum & product of roots of quadratic equation is gotten from the general expression of quadratic equation. If the distinct roots are α and β, then
α + β = -b/a (sum of roots)
αβ =c/a (product of roots)
Example 1 – find the sum and products of 2x2 + 3x – 1 = 0
Solution
2x2 + 3x – 1 = 0
a =2, b = 3, c = -1
Let α and β be the roots of the equation, then
α+β= -b/a= -3/2
αβ = c/a = -1/2
Example 2 – find the sum and products of 3x2 – 5x -2 = 0
Solution
3x2 -5x -2 =0
a= 3, b= -5, c= -2
let α and β be the roots of the equation, then
α+β= -b/a = 5/3
αβ= c/a= -2/3
NB: The quadratic equation whose root are α and β is
(X – α )(X – β) = 0
X2 – (α +β)X + αβ = 0
Example – Find the quadratic equation whose roots are 3 & -2
Solution
α=3 and β=-2
α+β = 3 + (-2) = 1
αβ = 3 x -2 = -6
X2 – (α +β)x + αβ = 0
X2 – (1)X + (-6) = 0
X2 – X -6 = 0
Symmetric Properties of Roots
Certain relations involving α and β can also be determined from α+β andαβ even when we do not know α and β distinctly. Such relations are usually said to be symmetric. They are symmetric in the sense that if α and α are interchanged, either the relation remains the same or is multiplied by -1
Example – If α ≠ β, determine whether or not each of the following is symmetric
(a) α+β (b) αβ (c) α2 + β2 (d) α2 – β2 (e) 3α +2β (f) α3 + β3
Solution
- α+β = β +α , therefore α+β is symmetric
- αβ =βα, therefore αβ is symmetric
- α2 + β2 = β2 + α2, therefore α2 + β2 is symmetric
- α2 – β2 = – (β2 – α2), thereforeα2 – β2 is symmetric
- 3α +2β ≠ 3β + 2α since α ≠ β, therefore 3α +2β is not symmetric
- α3 + β3 = β3 + α3, therefore α3 + β3 is symmetric
EVALUATION
- Find the quadratic equation whose roots are ½ and 5
- Find the sum & product of roots of the equation 3x2 – 5x – 2 = 0
ASSIGNMENT
- Solve for x ; x2 – 9 = 0 (a) 3, 3 (b) 3, -3 (c) 9, 3 (d) -3, -3
- Find the sum and products of 2x2 + 3x – 1 = 0 (a) -3/2, ½ (b) 3/2 , -1/2 (c) -3/2, -1/2 (d) 3, 2
- Solve for x : 6x2 – 13x + 5 = 0 (a) 5/3, ½ (b) -5/3,1/3 (c) 5/3, -1/3 (d) 3/5, 1/3
- Factorize 6x2 + 5x -6 (a) (3x – 2) (2x + 3) (b) (3x + 2) (2x + 3) (c)(3x – 2) (2x – 3) (d) (2x -3) (2x+ 2)
- Find the sum and products of x2 – 4x – 3 = 0
Theory
- Solve S = ut + ½ at2 using the completing the square method
- Find the quadratic equation whose roots are ¾ and ½