1. ST. WILLIAMS INSTITUTE
Magsingal, Ilocos Sur
JUNIOR HIGH SCHOOL DEPARTMENT
NAME: _____________________________________
GRADE & SECTION: ___________________
TEACHER: Mrs. Novy Reina B. Orteza
LEARNING MODULE IN
MATHEMATICS

2. LEARNING COMPETENCIES
At the end of the lesson, you are expected to:
1. solve quadratic equations by completing the square; and
using the quadratic formula
2. characterize the roots of a quadratic equation using the
discriminant.
3. describe the relationship between the coefficients and the
roots of a quadratic equation.
4. solve equations transformable to quadratic equations
(including rational algebraic equations).
5. solve problems involving quadratic equations and rational
algebraic equations.

3. LESSON PROPER

4. The expression x2 + bx becomes a perfect square when (
𝑏
2
)2
is added to it. Thus,
𝑥2
+ 𝑏𝑥 + (
𝑏
2
)
2
= (𝑥 +
𝑏
2
)
2
.
To solve a quadratic equation, this approach is called completing the square method.
In 𝑥2
+ 2𝑏𝑥, if we take 1/2 (2b) or b and square it, we would get the third term 𝑏2
.
Complete the square. Then write each completed square in factored form.
a.) 𝑥2
+ 10𝑥 b.) 𝑥2
− 7𝑥
SOLUTIONS:
a.) To complete the term, add (
10
2
)
2
or 25.
Completed square: 𝑥2
+ 10𝑥 + 5.
Factored form: (𝑥 + 5)2
.
Completing the Square
In an expression of the form 𝒙𝟐
+ 𝒃𝒙 or 𝒙𝟐
− 𝒃𝒙 , add the constant term (
𝑏
2
)
2
to
complete the square.

5. b.) To complete the term, add (−
7
2
)
2
or
49
4
.
Completed square: 𝑥2
− 7𝑥 +
49
4
.
Factored form: (𝑥 −
7
2
)
2
.
Complete the square. Then write each completed term in factored form.
1.) 𝑥2
+ 3𝑥 2.) 𝑥2
− 8𝑥 3.) 𝑥2
= 2𝑥

6. Solve the root of the quadratic equation 𝒙𝟐
+ 𝟏𝟖𝒙 = 0 by using the completing the
square method.
𝑥2
+ 18𝑥 = 0
𝑥2
+ 18𝑥 + 81 = 81 Divide 18 by 2 then square it.
(x + 9)2 = 81 Factor the left side of the equation.
(x +9) = ±9 Find the square root of both side of the equation.
x + 9 = 9 x + 9 = -9 Equate the factors to zero.
x = 9 -9 x = -9 – 9 Simplify.
x = 0 x = -18
Steps in Solving Quadratic Equations by Completing the Square Method
1. Transpose all terms containing the unknown to the left side of the equation
and the constant term to the right side, if necessary.
2. Divide each term of the equation by the numerical coefficient of the 𝑥2
term,
if necessary. This will change the equation to the form 𝑥2
+ 𝑏𝑥 = 𝑐.
3. Divide the coefficient of x by 2, square it, then add the result to both sides of
the equation.
4. Factor the left side of the equation. This is a perfect square trinomial. Simplify
the right side.
5. Take the square root of the expression on both sides of the equation. Write
the ± sign before the square root at the right side.
6. Equate the square root on the left side expression to the positive square root
on the right side in step 5. Solve for the first root. Then, find the second root of
the equation.
7. Equate the square root of the left side expression to the negative square root
of the right side in step 5. Solve for the second root.
8. Check each root by substituting it to the original equation.

7. Solve the roots of the following quadratic equations by using the completing the
square method. Show your pertinent solutions.
1.) 𝑥2
− 7𝑥 = 0 2.) 𝑥2
−
2
3
𝑥

8. THE DERIVATION OF THE QUADRATIC FORMULA
Therefore, any roots of the quadratic equation of the form 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, where 𝑎 ≠ 0
can be found by:
𝑥 =
− 𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
provided that the radicand, 𝑏2
− 4𝑎𝑐 ≥ 0.

9. If a quadratic equation is in the form 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, you can use the values for a,
b and c to find the solution of the equation. That is, you can find the values of x that will make
the equation true by using the quadratic formula.
Remember that the symbol ± is the easiest way to write the two solutions to the
equation. The formula actually means that:
𝒙 =
− 𝒃+ √𝒃𝟐−𝟒𝒂𝒄
𝟐𝒂
or 𝒙 =
− 𝒃− √𝒃𝟐−𝟒𝒂𝒄
𝟐𝒂
.
Unless the quantity 𝑏2
− 4𝑎𝑐 is negative or zero, the graph of 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
crosses the x-axis at exactly two points.
NOTE: The quadratic formula can be used to solve all quadratic equations including those for
which factoring method cannot be used.
The Quadratic Formula
For real numbers a, b, and c, with 𝒂 ≠ 𝟎, the solution of the quadratic
equation 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 is given by the quadratic formula:
𝒙 =
− 𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
Solving Quadratic Equations Using the Quadratic Formula
1. Write the equation in standard form (zero on one side of the equation.
2. List the numerical values of the coefficients a, b and c.
3. Write the quadratic formula.
4. Substitute the numerical values for a, b and c in the quadratic formula.
5. Simplify to get the exact solution.
6. Check the solution.

10. Solve the real roots of the equation 𝒙𝟐
− 𝟐𝒙 − 𝟏𝟗 = 𝟎 using the quadratic formula.
SOLUTION:
In x2 – 2x – 19 =0, a = 1 ; b = -2 ; c = -19.
𝒙 =
− 𝒃+ √𝒃𝟐−𝟒𝒂𝒄
𝟐𝒂
𝒙 =
− (−𝟐)+ √(−𝟐)𝟐−𝟒(𝟏)(−𝟏𝟗)
𝟐(𝟏)
Substitute the values of a, b and c.
𝒙 =
𝟐+ √𝟒+𝟕𝟔
𝟐
Simplify.
𝒙 =
𝟐+ √𝟕𝟖
𝟐
Since 78 does not have any perfect square factor, therefore, the roots are 𝒙 =
𝟐+ √𝟕𝟖
𝟐
.
Solve the roots of the followingequationsby using the quadratic formula. Show your
pertinent solutions.
𝑥2
+ 4𝑥 − 5 = 0 2𝑥2
− 3x + 3 = 0 𝑥2
− 3 = 2𝑥

11. By inspection, determine the nature of the roots of the following equation.
a.) 𝑥2
− 7𝑥 − 8 = 0
The equation gives a = 1, b = -7 and c = -8. Substituting the given values,
𝑏2
− 4𝑎𝑐 = (−7)2
− 4(1)(−8)= 49 +32 = 81
Since 81 is greater than 0 and a perfect square, then the equation has 2 real roots.
Discriminant
The radicand 𝑏2
− 4𝑎𝑐 in the quadratic formula is called the discriminant.
Using the Discriminant
Given the quadratic equation in the form 𝑎𝑥2
+ 𝑏𝑥 + [ 𝑐 = 0, where a, b
and c are real numbers and 𝑎 ≠ 0, we can determine the number and type of
solutions of a quadratic equation by evaluating the discriminant 𝑏2
− 4𝑎𝑐.
 If 𝒃𝟐
− 𝟒𝒂𝒄 > 𝟎, the equation has two real solutions. Both will be rational
if the discriminant is a perfect square or irrational, if otherwise.
 If 𝒃𝟐
− 𝟒𝒂𝒄 = 𝟎, the equation has only one real solution which will be a
rational number.
 If 𝒃𝟐
− 𝟒𝒂𝒄 < 𝟎, the equation has no real number solution.
1. The sum of the roots is the additive inverse of the quotient of b and a.
r1 + r2 = −
𝒃
𝒂
2. The product of the roots is the quotient of c and a.
r1 × r2 =
𝒄
𝒂
.

12. Without solving for the solutions of the quadratic equation4𝑥2
+ 2𝑥 − 5 = 0, find the
sum and product of its roots.
SOLUTION: Given the equation4𝑥2
+ 2𝑥 − 5 = 0, we have a=4, b= 2 and c = -5.
Hence,
the sum of the roots is −
𝑏
𝑎
= −
2
4
= −
1
2
.
the product of the roots is
𝑐
𝑎
= −
5
4
.
Complete the following table by using the discriminant.
EQUATION 𝒃𝟐
− 𝟒𝒂𝒄
𝒃
𝒂
𝒄
𝒂
ROOTS r1 + r2 r1 × r2
r1 r2
𝟐𝒙𝟐
− 𝟓𝒙 + 𝟏 = 𝟎
𝟒𝒙𝟐
+ 𝟔𝒙 + 𝟑 = 𝟎
𝟐𝒙𝟐
+ 𝟏𝟏𝒙 − 𝟏𝟓 = 𝟎
𝟒𝒙𝟐
+ 𝟑𝒙 + 𝟖 = 𝟎
𝟔𝒙𝟐
− 𝟓𝒙 + 𝟖 = 𝟎
To derive the quadratic equation when the two roots are given, subtract each
root from x to get the corresponding linear factors and equate the product of the linear
factors to zero.

13. A quadratic equation is not always in the form 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0. If a quadratic is not
in this form, we can perform whatever operations are necessary to transform it to this form.
Some possible operations are any of the axioms or properties we have developed for
transforming equivalent fractions.
A partial list would include the following:
1. Clearing of fractions by multiplying all terms by the LCD. (Note: If the LCD includes the
variable, check all apparent solutions to be sure that no denominators become zero.)
2. Removing the parenthesis and other symbols.
3. Removing the radical signs by squaring both members of the equations.
4. Collecting like terms.
A quadratic system involves at least one quadratic equations. A quadratic
system with at least one linear equation is called a quadratic-linear system. A
quadratic system with two quadratic equations is called a quadratic-quadratic
system.