Second Term Lesson Note for Week Two
Class : SSS 1
Subject : Mathematics
Topic : Quadratic Equations formula
Duration : 80 Minutes
Period : Double Period
Reference Book :
- Essential Mathematics for Senior Secondary Schools, Book 1
- Lagos State Unified schemes of work for Senior Secondary Schools, SSS 1 – 3.
- Online Resources
Instructional Material : Chart showing the quadratic equation formula.
Learning Objectives : By the end of the lesson the student will be able to :
i. Derivative of the Roots of the General Form of Quadratic Equation.
ii. Using the Formula Methods to solve Quadratic Equations
iii. Sum and Product of quadratic roots.
Content :
General form of quadratic equation leading to Formular method
Derivative of the Roots of the General Form of Quadratic Equation
The general form of a quadratic equation is ax² + bx + c = 0. The roots of the general equation are found by completing the square.
ax² + bx + c = 0
Divide through by the coefficient of x2.
(ax²) /a + (bx) / a + c /a = 0
x² + bx/a + c/a = 0
x² + bx/a = 0 – c/a
x² + bx/a = – c/a
The square of half of the coefficient of x is
½ x b 2 = b 2
a 2a
Add b² /2a to both sides of the equation.
2a
x2 + (bx)\2a + (b²)/ 2a = – C/2a + (b²)/2a
= – C+ b2
a 4a2
x + b² = – 4ac + b²
2a 4a2
i.e x +b 2 = b2 – 4ac
2a 4a2
Take square roots of both sides of the equation :
=
i.e x + b= ± √ b² – 4ac
2a 2a
x =
Hence,
x = (-b ±√ b² – 4ac) /2a
EVALUATION
Suppose the general quadratic equation is Dy² + Ey + F = 0
Using the method of completing the square, derive the roots of this equation
Using the Formula Methods to Solve Quadratic Equations
Examples
Use the formula method to solve the following equations. Give the roots correct to 2 decimal places :
i.) 3x² – 5x – 3 = 0
ii.) 6x² + 13x + 6 = 0
iii.) 3x² – 12x + 10 = 0
Solution :
1. 3x² – 5x – 3 = 0
Comparing 3x² – 5x – 3 = 0
with ax² + bx + C = 0
a = 3, b = -5, C = -3
Since
X = [-b ±√b2 – 4ac] / 2a
x = [-(-5) ±√ (-5)² – 4 x 3 x -3] / (2×3)
x = [+ 5 ± √ 25 + 36] / 6
x = [+ 5 ± √61] / 6
x = [+ 5 + 7.810] / 6 = + 12.810 / 6
or
x = [+5 – 7.810] / 6 = -2.810 / 6
x = 12.810 / 6 or x = – 2.810 / 6
i.e.x = 2.135 or x = – 0.468
x = 2.14 or x = – 0.47 (to 2 decimal places)
2.) 6x² + 13x + 6 = 0
comparing 6x² + 13x + 6 = 0
with ax² +bx + c = 0
a = 6, b = 13, c = 6
Since
x = [-b ± √ b2 – 4ac] / 2a
2a
x = [- 13 ±√ (13)² – 4 x 6 x 6] / (2 x 6)
x = [-13 ±√169 – 144] / 12
x = [- 13 ±√25] / 12
x = = [-13 ± 5] / 12
12
x = [-13 + 5] / 12 or x = [-13 – 5] / 12
x = -8 /12 or x = – 18 /12
x= -2/3 or x = -3/2
x=- 0.666 or x = – 1.50
i.e x= 0.67 or x = -1.50 to 2 decimal places
(3) 3x² – 12x + 10 = 0
Comparing 3x² – 12x + 10 = 0 with the equation ax² +bx + c = 0, then,
a = 3, b= -12, c = 10.
Since x = [ – b ± √b² – 4ac] / 2a then
x =[ – (-12) ±√(-12)² – 4 x 3 x 10] / (2×3)
x = [+ 12 ±√ 144 – 120] / 6
x = [+ 12 ±√24] /6
x = [12 ± 4.899] / 6
x = [+ 12 + 4.899] / 6 = 16.899 / 6
or x = [ + 12 – 4.899] / 6 = 7.101 / 6
i.e x = 16.899/ 6 or x = 7.101/ 6
x = 2.8165 or x = 1.1835
i.e . x = 2.82 or x = 1.18 (to 2 decimal places) .
EVALUATION
Use the formula method to solve the following quadratic equations :
1. t² – 8t + 2 = 0
2. t² + 3t + 1 = 0
Sum and Product of quadratic roots.
We can find the sum and product of the roots directly from the coefficient in the equation
It is usual to call the roots of the equation α and β If the equation
ax² +bx + c = 0 ……………. I
has the roots α and β then it is equivalent to the equation
(x – α )( x – β ) = 0 ………..………… 2
Divide equation (1) by the coefficient of x²
(ax²) /a + (bx) /a + c / a = 0 ………………… 3
Comparing equations (2) and (3)
x² + bx /a+ c/a = 0
x2 – ( α +β)x + αβ = 0
then
α+ β= -b/a and αβ = c/a
For any quadratic equation,
ax² +bx + c = 0 with roots α and β
α + β = -b/a and αβ = c/a
Examples
1. If the roots of 3x² – 4x – 1 = 0 are α and β, find α + β and αβ
2. If α and β are the roots of the equation
3x² – 4x – 1 = 0 , find the value of
(ai) α + β (aii) β α
(b) α – β
Solutions
1a. Since α + β = -b/ a
Comparing the given equation 3x² – 4x – 1= 0 with the general form
ax² + bx + c = 0
a = 3, b = – 4, c = – 1.
Then,
α + β = -b/a = – (- 4)/3
= + 4 / 3
= +1 1/3
αβ = c/a = -1/3
= -1/3
2. (a) α + β = α² +β² + 2αβ
= (α + β )² – 2αβ
αβ
Here, comparing the given equation, with the general equation,
a = 3, b = – 4, c = – 1
from the solution of example 1 (since the given equation are the same ),
α + β = -b/a= – (-4) / 3
= +4/ 3
αβ = c / a = – 1/ 3
then
α + β = ( α+ β )² – 2 αβ
= (4/3 )² – 2 (- 1/3)
-1/3
= 16/9 + 2/3
– 1/3
= (16 + 6 )/ 9÷ -1/3
22 x -3
9 1
= -22
3
or α + β = – 22 = – 7 1/3
β α 3
(b) Since
(α-β) 2 =α2 + β 2-2α β
but
α2 + β2 = (α + β)2 -2 α β
:.(α- β)2 = (α+ β)2 – 2αβ -2αβ
(α – β)2 = (α + β)2 – 4α β
:.(α – β) = √(α + β )2 – 4αβ
(α – β) =√(4/3)2 – 4 (- 1/3)
= √16/9 +4/3
=
= =
:. α – β = √28
3
EVALUATION
If α and β are the roots of the equation 2x² – 11x + 5 = 0, find the value of α – β
GENERAL EVALUATION
Solve the following quadratic equations:
1. 63z = 49 + 18z²
2. 8s² + 14s = 15
Solve the following using formula method :
1.) 12y² + y – 35 = 0
2.) h² – 15h + 54 = 0
READING ASSIGNMENT
New General Mathematics SS Bk2 pages 41-42 ,Ex 3e Nos 19 and 20 page 42.
WEEKEND ASSIGNMENT
If α and β are the roots of the equation 2×2 – 7x – 3 = 0 find the value of:
i. α + β (a) 2/3 (b) 7/2 (c) 2/5 (d) 5/3
ii. αβ (a) -3/2 (b) 2/3 (c) 3/2 (d) – 2/3
iii. αβ² + α²β (a) 21/4 (b) 4/21 (c) – 4/21 (d) -21/4
Solve the following equation using the formula method.
1. 6p² – 2p – 7 = 0
2. 3 = 8q – 2q².
THEORY
1. Solve the equation 2x² + 6x + 1 = 0 using the formula method
2. If α and β are the roots of the equation 3x² – 9x + 2 = 0, find the values of
i. αβ² + α²β
ii. α² – αβ + β²