The document provides 3 examples of solving quadratic equations by setting them equal to zero and using the quadratic formula. Each example shows the step-by-step work of isolating the constant term, factoring the equation, taking the square root of both sides to solve for the roots, and checking the solutions. The examples demonstrate how to solve quadratic equations from setting them equal to zero through finding the solution set.
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Strategic Intervention Material (SIM) was provided for Grade 10 students to enhance learning and to motivate and stir up the attention and interest of the students until they master the topic. This material depicts the entire definition of learning since it concludes a systematic development of students’ comprehension on a distinct lesson in Mathematics 10.
Quadratic Equations
In One Variable
1. Quadratic Equation
an equation of the form
ax2 + bx + c = 0
where a, b, and c are real numbers
2.Types of Quadratic Equations
Complete Quadratic
3x2 + 5x + 6 = 0
Incomplete/Pure Quadratic Equation
3x2 - 6 = 0
3.Solving an Incomplete Quadratic
4.Example 1. Solve: x2 – 4 = 0
Solution:
x2 – 4 = 0
x2 = 4
√x² = √4
x = ± 2
5.Example 2. Solve: 5x² - 11 = 49
Solution:
5x² - 11 = 49
5x² = 49 + 11
5x² = 60
x² = 12
x = ±√12
x = ±2√3
6.Solving Quadratic Equation
7.By Factoring
Place all terms in the left member of the equation, so that the right member is zero.
Factor the left member.
Set each factor that contains the unknown equal to zero.
Solve each of the simple equations thus formed.
Check the answers by substituting them in the original equation.
8.Example: x² = 6x - 8
Solution:
x² = 6x – 8
x² - 6x + 8 = 0
(x – 4)(x – 2) = 0
x – 4 = 0 | x – 2 = 0
x = 4 x = 2
9.By Completing the Square
Write the equation with the variable terms in the left member and the constant term in the right member.
If the coefficient of x² is not 1, divide every term by this coefficient so as to make the coefficient of x² equal to 1.
Take one-half the coefficient of x, square this quantity, and add the result to both members.
Find the square root of both members, placing a ± sign before the square root of the right member.
Solve the resulting equation for x.
10.Example: x² - 8x + 7 = 0
11.By Quadratic Formula
Example: 3x² - 2x - 7 = 0
12.Solve the following:
1. x² - 15x – 56 = 0
2. 7x² = 2x + 6
3. 9x² - 3x + 8 = 0
4. 8x² + 9x -144 = 0
5. 2x² - 3 + 12x
13.Activity:
Solve the following quadratic formula.
By Factoring By Quadratic Formula
1. x² - 5x + 6 = 0 1. x² - 7x + 6 = 0
2. 3 x² = x + 2 2. 10 x² - 13x – 3 = 0
3. 2 x² - 11x + 12 = 0 3. x (5x – 4) = 2
By Completing the Square
1. x² + 6x + 5 = 0
2. x² - 8x + 3 = 0
3. 2 x² + 3x – 5 = 0
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2. Example 1. Solve 2x2 + 4x – 16 = 0
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
2x2 + 4x = 16
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 2.
2x2
2
+
4𝑥
2
=
16
2
x2 + 2x = 8
3. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 + 2x = 8
2
2
= 1 ; (1)2 = 1
x2 + 2x + 1 = 8 + 1
x2 + 2x + 1 = 9
(x + 1)2 = 9
4. Apply the Square Root Property.
(Square both sides)
x + 1)2 = 9
x + 1 = ±3
5. Solve for the roots. x + 1 = ±3
x + 1 = 3 and x + 1 = -3
x = 3 - 1 and x = -3 - 1
x = 2 and x = -4
4. Example 2. Solve for the roots of 2x2 – 8x = 120
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
2x2 – 8x = 120
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 2.
2x2
2
-
8𝑥
2
=
120
2
x2 – 4x = 60
5. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 – 4x = 60
=
−4
2
= -2
= (-2)2 = 4
x2 – 4x + 4 = 60 + 4
x2 – 4x + 4 = 64
(x – 2)2 = 64
4. Apply the Square Root Property.
(Square both sides)
(x – 2)2 = 64
x - 2 = ±8
5. Solve for the roots. x – 2 = 8 and x – 2 = -8
x = 8 + 2 and x = -8 + 2
x = 10 and x = -6
6. Example 3. Find the solution set of 3x2 + 48 = -30x.
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
3x2 + 30x = -48
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 3.
3x2
3
+
30𝑥
3
=
−48
3
x2 + 10x = -16
7. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 + 10x = -16
10
2
= 5
(5)2 = 25
x2 + 10x + 25 = -16 + 25
(x + 5)2 = 9
4. Apply the Square Root Property.
(Square both sides)
(x + 5)2 = 9
x + 5 = ±3
5. Solve for the roots. x + 5 = 3 and x + 5 = -3
x = 3 – 5 and x = -3 – 5
x = -2 and x = -8
