THE LIPA GRACE ACADEMY INC.
HIGH SCHOOL DEPARTMENT
#556 Interior B. Morada Avenue, Lipa City
S.Y. 2024-2025
S.Y. 2024-202
MATHEMATICS
GRADE 9
❖Solving Quadratic
Equations by Using
the Quadratic
Formula
❖
Discriminant
and Nature of Roots
of Quadratic
Equation
Solving Quadratic Equations by Using
the Quadratic Formula
❖
❖ Reviewing…
Activity 1: It’s Good to Be Simple!
Directions: simplify the following expressions.
1.
2.
3.
4.
5.
6+ √9
Processing questions:
a. How would you describe the
expressions given?
b. How did you simplify each
expression?
2(3)
−9− √24
2(2)
−6+ √18
2(2)
5+ √25+100
2(4)
−4− √42 +16
2(4)
The solutions of any quadratic equation ax2 + bx
+ c = 0 can be determined using the quadratic
formula x=
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
, a≠ 0.
To use the Quadratic Formula, we substitute the
values of a, b, and c into the expression on the
right side of the formula. Then, we do all the
math to simplify the expression. The result gives
the solution(s) to the quadratic equation.
➢ To solve any quadratic equation ax2 + bx + c = 0 using the quadratic formula,
1. Determine the values of a, b, and c
2. Substitute these in the equation x=
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
.
3. Simplify the results, if possible.
4. then check the solutions obtained against the original equation.
Examples:
❖ Solve the following quadratic equation using the quadratic formula.
a. 2x2 + 3x = 27
b. x2+10x+9=0
c. x2-11x+24=0
d. 12x2+5x=3
Solutions:
a. 2x2 + 3x = 27
a= 2, b= 3, c= -27
STEP 1
x=
x=
STEP 2
x=
x=
x=
STEP 3
x=
−𝑏± √𝑏2−4𝑎𝑐
2𝑎
2 −4(2)(−27)
√3
−3±
2(2)
−3± √9−(−216)
4
−3± √9+216
4
−3± √225
STEP 4
Checking: x=3
2x2 + 3x = 27
2(3)2 + 3(3) = 27
2(9) + 3(3) = 27
2(9) + 9 = 27
18 + 9 = 27
27 = 27
−9
Checking: x= 2
2x2 + 3x = 27
−9
−9
2( 2 )2 + 3( 2 ) = 27
4
−3± 15
4
81
x=
−3+ 15
x=
12
x=
4
x=
4
x= 3
x=
−9
−3− 15
2( 4 ) + 3( 2 ) = 27
4
−18
( 2 ) + ( 2 ) = 27
4
−9
2
2
81
54
−27
= 27
27 = 27
The roots/solutions of the quadratic equation 2x2 + 3x = 27 are 3 and
−9
2
b. x2+10x+9=0
a= 1,
STEP 1
x=
x=
STEP 2
x=
x=
STEP 3
x=
x=
−10+ 8
x=
−2
2
2
x=−1
b= 10,
STEP 4
c= 9
Checking: x= -1
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
−10± √102 −4(1)(9)
x2+10x+9=0
(-1)2+10(-1)+9=0
1 + (-10) +9 = 0
-9 + 9 = 0
0= 0
2(1)
−10± √100−36
2
−10± √64
2
−10± 8
Checking: x=−9
2
x=
x=
−10− 8
2
−18
2
x2+10x+9=0
(-9)2+10(-9)+9=0
81 + (-90) +9 = 0
-9 + 9 = 0
0= 0
x=−9
The roots/solutions of the quadratic equation x2+10x+9=0 are -1 and −9.
c. x2-11x+24=0
a= 1,
STEP 1
x=
x=
STEP 2
x=
x=
x=
x=
11+ 5
x=
16
b= -11,
c= 24
STEP 4
Checking: x= 3
x2-11x+24=0
(3)2-11(3)+24=0
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
−(−11)± √(−11)2 −4(1)(24)
9 - 33 +24 = 0
-24 + 24 = 0
0= 0
2(1)
11± √121−96
2
11± √25
Checking: x=8
2
11± 5
x2-11x+24=0
(8)2-11(8)+24=0
2
2
2
x=8
x=
11− 5
x=
6
64 -88 + 24 = 0
-24 + 24 = 0
0= 0
2
2
x=3
The roots/solutions of the quadratic equation x2-11x+24=0 are 3 and 8.
d. 12x2+5x=3
a= 12, b= 5,
STEP 1
x=
x=
STEP 2
x=
x=
STEP 3
x=
x=
−5+ 13
x=
8
x=
24
1
3
24
c=-3
STEP 4
1
3
Checking: x=
12x2+5x=3
1
1
3
3
1
5
12( ) + ( ) = 27
9
3
12
5
( )+ = 27
9
3
4
5
+ = 27
3
3
9
=3
3
12( )2 + 5( ) = 3
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
−5± √(5)2 −4(12)(−3)
2(12)
−5± √25+144
24
−5± √169
3=3
24
−5± 13
24
Checking: x=
x=
x=
x=
−3
4
−5−13
12x2+5x=3
24
−18
12(
24
−3
12( ) + (
4
−3 2
−3
) + 5( ) = 3
4
4
9
−15
)=3
16
4
108
−15
( )+
= 27
16
4
27
−15
+
= 27
4
4
12
4
=3
3=3
1
−3
The roots/solutions of the quadratic equation 12x2+5x=3 are 3 and 4 .
Activity 2:
a. x2−9𝑥 =−10
b. 2𝑚2+13𝑚+20 =0
c. 2𝑦2+8𝑦 =9
Discriminant and Nature of Roots of
Quadratic Equations…
Recalling Basic Concept:
Activity 1: Classification Bin!
Directions: The image is a classroom hidden number game. In the game you must find out the
hidden numbers and place it in the appropriate trash bin.
Reminder: a number can be written in two trash bins.
Processing Question:
1. How did you categorize each number on the hidden number game?
We use the formula 𝑥 =
2
ax + bx + c = 0 where a ≠ 0.
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
for the roots of the quadratic equation of the form
The expression under the radical sign of the quadratic formula plays an important role in
the calculation of the roots. This expression enables us to determine the discriminant and
nature of roots without solving the equation. By the nature of roots we mean:
• whether the equation has real roots.
• if there are real roots, whether they are different or equal.
The expression b2 – 4ac is called the discriminant of the quadratic equation because it
discriminates among the four cases which can occur.
•
If D = 0, then the roots are real, rational, equal.
•
If D > 0 and is a perfect square, then the roots are real, rational, and unequal.
•
If D > 0 and is not a perfect square, then the roots are real, irrational, and unequal.
•
If D < 0, then the roots are non-real/imaginary or complex.
Examples:
Determine the discriminant and nature of roots of each quadratic equation.
1. x2 – 6x + 9 = 0
Solution:
a=1, b=-6, c=9
D= b2 – 4ac
D= (-6)2 – 4(1)(9)
D= 36-36
D= 0
2. x2 – 4x + 3 = 0
Solution:
a=1, b=-4, c=3
D= b2 – 4ac
D= (-4)2 – 4(1)(3)
D= 16-12
D= 4
3. x2 – 7x – 4 = 0
Solution:
a=1, b=-7, c=-4
D= b2 – 4ac
D= (-7)2 – 4(1)(-4)
D= 49+16
D= 65
Answer:
D= 0
✓ Real, rational and equal roots.
Answer:
D= 4> 0 and a perfect square
✓ Real, rational and unequal roots.
Answer:
D= 65> 0 and not a perfect square
✓ Real, irrational and unequal roots.
4. 2x2 - 3x + 5 = 0
Solution:
a=2, b=3, c=5
D= b2 – 4ac
D= (-3)2 – 4(2)(5)
D= 9-40
D= -31
Answer:
D= -31< 0 and complex
✓ Imaginary and complex roots.
❖ The Principle of General Equation
•
The sum of the roots of a quadratic equation is the additive of the quotient of b and a.
𝑏
•
𝑟1 + 𝑟2 = − 𝑎
The product of the roots of a quadratic equation is the quotient of c and a.
𝑐
𝑟1 ∙ 𝑟2 = 𝑎
Examples:
1. 𝒙𝟐 + 𝟑𝒙 − 𝟐𝟖 = 𝟎
𝑏
−3
SUM: 𝑟1 + 𝑟2 = − 𝑎 = 1 = −3
𝑐
PRODUCT: 𝑟1 ∙ 𝑟2 = 𝑎 =
−28
1
= −28
2. 𝒙𝟐 − 𝟏𝟐𝒙 + 𝟑𝟓 = 𝟎
𝑏
SUM: 𝑟1 + 𝑟2 = − 𝑎 =
−(−12)
𝑐
1
35
= 12
PRODUCT:𝑟1 ∙ 𝑟2 = 𝑎 = 1 = 35
Example:
• What is the quadratic equation whose roots are -5 and 7?
Method 1: Use sum and product of the roots. Then, use the formula.
𝑥 2 − (𝑟1 + 𝑟2 )𝑥 + 𝑟1 ∙ 𝑟2 = 0
Method 2: Use (𝑥 − 𝑟1 ) (𝑥 − 𝑟2 ) = 0
Solution:
Method 1: Use sum and product of the roots.
𝑟1 + 𝑟2 = −5 + 7 = 2
𝑟1 ∙ 𝑟2 = −5 ∙ 7 = −35
Then, use the formula.
𝑥 2 − (𝑟1 + 𝑟2 )𝑥 + 𝑟1 ∙ 𝑟2 = 0
𝑥 2 − 2𝑥 − 35 = 0
Method 2: Use (𝑥 − 𝑟1 ) (𝑥 − 𝑟2 ) = 0
(𝑥 − (−5)) (𝑥 − 7) = 0
(𝑥 + 5) (𝑥 − 7) = 0
Use FOIL Method:
F
𝑥 ∙ 𝑥 = 𝑥2
O
𝑥 ∙ −7 = −7𝑥
I
5 ∙ 𝑥 = 5𝑥
L
5 ∙ −7 = −35
Answer:
𝑥 2 − 2𝑥 − 35 = 0
−7𝑥 + 5𝑥 = −2𝑥
Activity 2: “Roots to Describe "
Find and analyze the roots of each equation provided. Let r1 and r2 be the roots of the
equation. Complete the table below by placing a checkmark in the cells that correspond to the
characteristics of the roots you have determined for each equation.
Unequal
EQUAL OR
UNEQUAL? The value
of b2-4ac/
Discrimi
nant
Equal
Irrational
r2
RATIONAL OR
IRRATIONAL?
Rational
r1
Imaginary
EQUATIONS
REAL OR
IMAGINARY?
Real
ROOTS
x2-7x+10=0
x2-4x+7=0
x2-10x+25=0
x2-4x-2=0
Processing Question:
1. What are the implications if the discriminant is zero versus when it is positive or
negative?
2. How do you interpret the discriminant when it is positive, zero, or negative?
MINI PETA #1: SOLVE MY TRAITS
Directions: Follow the instructions below and write your answers on a whole sheet of paper.
1. Solve the Quadratic Equations:
o Solve the following quadratic equations and show your work:
1. x2−4x−5=0
2. x2+2x−8=0
3. x2−6x+9=0
2. Calculate the Discriminants:
o For each equation, calculate the discriminant using D=b 2−4ac.
o Write down the value of the discriminant for each equation.
3. Interpret the Results:
o Explain what the discriminant value tells you about the nature of the roots
(real and distinct, real and repeated, or complex) for each equation. Write 1-2
sentences for each.
Standards for Scoring: