The document presents a comprehensive lesson on solving quadratic equations, including those not in standard form and rational algebraic equations transformable to quadratic equations. It provides multiple examples demonstrating how to transform equations into the standard form, solutions using various methods (factoring, quadratic formula, etc.), and outlines steps for solving rational equations. Each lesson is accompanied by detailed examples with solutions, culminating in solution sets for each problem.

1. MATHEMATICS 9
Solving Equations Transformable to Quadratic
Equation Including Rational Algebraic
Equations
Lesson 1: Solving Quadratic Equations That Are Not
Written In Standard Form
Lesson 2: Solving Rational Algebraic Equations
Transformable To Quadratic Equations

2. In solving quadratic equation that is not written in standard form,
transform the equation in the standard form ax2 + bx + c = 0 where a,
b, and c are real numbers and a ≠ 0 and then, solve the equation
using any method in solving quadratic equation (extracting square
roots, factoring, completing the square, or quadratic formula).
Lesson 1: Solving Quadratic Equations That
Are Not Written In Standard Form

3. Example 1: Solve x(x – 5) = 36.
Solution:
Transform the equation in standard form.
x(x – 5) = 36
x2 – 5x = 36
x2 – 5x – 36 = 0
Solve the equation using any method.
By factoring
x2 – 5x – 36 = 0 x – 9 = 0 x + 4 = 0
(x – 9)(x + 4) = 0 x = 9 x = – 4
• The solution set of the equation is {9, – 4}.

4. Example 2: Solve (x + 5)2 + (x – 2)2 = 37.
Solution:
Transform the equation in standard form.
(x + 5)2 + (x – 2)2 = 37
x2 + 10x + 25 + x2 – 4x + 4 = 37
2x2 + 6x + 29 = 37 → 2x2 + 6x + 29 – 37 = 0
2x2 + 6x – 8 = 0
x2 + 3x – 4 = 0 Divide all terms by 2.
Solve the equation using any method.
By factoring
x2 + 3x – 4 = 0 x + 4 = 0 x – 1 = 0
(x + 4)(x – 1) = 0 x = – 4 x = 1
• The solution set of the equation is {– 4, 1}.

5. Example 3: Solve 2x2 – 5x = x2 + 14.
Solution:
Transform the equation in standard form.
2x2 – 5x = x2 + 14
2x2 – x2 – 5x – 14 = 0
x2 – 5x – 14 = 0
Solve the equation using any method.
By Quadratic Formula, identify the values of a, b, and c
a = 1, b = -5, c = -14
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
=
−(−5) ± (−5)2 −4(1)(−14)
2(1)
=
5 ± 25 + 56
2
=
5 ± 81
2
=
5 ± 9
2
𝑥 =
5+9
2
=
14
2
= 𝟕 𝑥 =
5−9
2
=
−4
2
= −𝟐
The solution set of the equation is {– 2, 7}.

6. Example 4: Solve (x – 4)2 = 4.
Solution:
Transform the equation in standard form.
(x – 4)2 = 4
x2 – 8x + 16 = 4
x2 – 8x + 16 – 4 = 0
x2 – 8x + 12 = 0
Solve the equation using any method.
By factoring
x2 – 8x + 12 = 0 x – 6 = 0 x – 2 = 0
(x – 6)(x – 2) = 0 x = 6 x = 2
• The solution set of the equation is {6, 2}.

7. Example 5: Solve (3x + 4)2 – (x – 1)2 = – 5.
Solution:
Transform the equation in standard form.
(3x + 4)2 – (x – 1)2 = – 5
9x2 + 24x + 16 – (x2 – 2x + 1) = – 5
9x2 + 24x + 16 – x2 + 2x – 1 = – 5
8x2 + 26x + 15 = – 5 → 8x2 + 26x + 15 + 5 = 0
8x2 + 26x + 20 = 0
4x2 + 13x + 10 = 0 Divide all terms by 2.
Solve the equation using any method.
By Quadratic Formula, identify the values of a, b, and c
a = 4, b = 13, c = 10
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
=
−(13) ± (13)2 −4(4)(10)
2(4)
=
−13 ± 169 − 160
8
=
−13 ± 9
8
=
−13 ± 3
8
𝑥 =
−13+3
8
=
−10
8
= −
𝟓
𝟒
𝑥 =
−13−3
8
=
−16
8
= −𝟐
• The solution set of the equation is {– 2, −
𝟓
𝟒
}.

8. Lesson 2: Solving Rational Algebraic Equations
Transformable To Quadratic Equations
There are rational equations that can be transformed
into quadratic equation of the form ax2 + bx + c = 0 where a,
b and c are real numbers, and a ≠ 0 and it can be solved
using the different methods in solving quadratic equation.

9. Steps in Solving Rational Equations:
1. Multiply both sides of the equation by the Least Common
Multiple (LCM) or Least Common Denominator (LCD).
2. Write the resulting quadratic equation in standard form.
3. Solve the equation using any method in solving quadratic
equation.
4. Check whether the obtained values of x satisfies the given
equation.

10. Example 1: Solve the rational algebraic equation
6
𝑥
+
𝑥−3
4
= 2.
Solution:
1. Multiply both side of the equation by the LCD, the LCD is 4x.
4𝑥
6
𝑥
+
𝑥−3
4
= 4𝑥(2) → 4(6) + x(x – 3) = 8x
24 + x2 – 3x = 8x
2. Transform the resulting equation in standard form.
24 + x2 – 3x = 8x → x2 – 3x – 8x + 24 = 0
x2 – 11x + 24 = 0
3. Solve the equation using any method. Since the equation is factorable,
x2 – 11x + 24 = 0 x – 3 = 0 x – 8 = 0
(x – 3)(x – 8) = 0 x = 3 x = 8
The solution set of the equation is {3, 8}.

11. Example 2: Solve the rational algebraic equation
1
𝑥
+
1
𝑥+1
=
7
12
.
Solution:
1. Multiply both side of the equation by the LCD, the LCD is 12x(x + 1).
12𝑥(𝑥 + 1)
1
𝑥
+
1
𝑥+1
= 12𝑥 𝑥 + 1
7
12
→ 12(x + 1) + 12x = x(x+1)(7)
12x + 12 + 12x = 7x2 + 7x
2. Transform the resulting equation in standard form.
12x + 12 + 12x = 7x2 + 7x → 0 = 7x2 + 7x – 12x – 12x – 12
0 = 7x2 – 17x – 12
7x2 – 17x – 12 = 0
3. Solve the equation using any method. By Quadratic Formula, identify the values of a, b, and c. a = 7, b = – 17, c = – 12
𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
=
−(−17)± (−17)2 −4(7)(−12)
2(7)
=
17± 289+336
14
=
17± 625
14
=
17±25
14
𝑥 =
17+25
14
=
42
14
= 𝟑 𝑥 =
17−25
14
=
−8
14
= −
𝟒
𝟕
• The solution set of the equation is {3, −
𝟒
𝟕
}.

12. Example 3: Solve the rational algebraic equation 𝑥 +
8
𝑥−2
= 1 +
4𝑥
𝑥−2
.
Solution:
1. Multiply both side of the equation by the LCD, the LCD is x – 2.
𝑥 − 2 𝑥 +
8
𝑥−2
= 𝑥 − 2 1 +
4𝑥
𝑥−2
→ x(x – 2) + 8 = 1(x – 2) + 4x
x2 – 2x + 8 = x – 2 + 4x
x2 – 2x + 8 = 5x – 2
2. Transform the resulting equation in standard form.
x2 – 2x + 8 = 5x – 2 → x2 – 2x – 5x + 8 + 2 = 0
x2 – 7x + 10 = 0
3. Solve the equation using any method. Since the equation is factorable,
x2 – 7x + 10 = 0 x – 5 = 0 x – 2 = 0
(x – 5)(x – 2) = 0 x = 5 x = 2
• The solution set of the equation is {5, 2}.

13. Example 4: Solve the rational algebraic equation
𝑥+3
3
+
1
𝑥−3
= 4.
Solution:
1. Multiply both side of the equation by the LCD, the LCD is 3(x – 3).
3(𝑥 − 3)
𝑥+3
3
+
1
𝑥−3
= 3(𝑥 − 3)(4) → (x – 3)(x + 3) + 3(1) = 12(x – 3)
x2 – 9 + 3 = 12x – 36
2. Transform the resulting equation in standard form.
x2 – 6 = 12x – 36 → x2 – 12x – 6 + 36 = 0
x2 – 12x + 30 = 0
3. Solve the equation using any method. By Quadratic Formula, identify the values of a, b, and c. a = 1, b = – 12, c = 30
𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
=
−(−12)± (−12)2 −4(1)(30)
2(1)
=
12± 144−120
2
=
12± 24
2
=
12±2 6
2
𝑥 =
12+2 6
2
= 𝟔 + 𝟔 𝑥 =
12−2 6
2
= 𝟔 − 𝟔
• The solution set of the equation is 𝟔 + 𝟔 , 𝟔 − 𝟔 .

14. Example 4: Solve the rational algebraic equation
𝑥
3𝑥+2
=
2
𝑥+1
.
Solution:
1. Multiply both side of the equation by the LCD, the LCD is (3x + 2)(x + 1).
(3𝑥 + 2)(𝑥 + 1)
𝑥
3𝑥+2
= (3𝑥 + 2)(𝑥 + 1)
2
𝑥+1
→ x(x + 1) = 2(3x + 2)
x2 + x = 6x + 4
2. Transform the resulting equation in standard form.
x2 + x = 6x + 4 → x2 + x – 6x – 4 = 0
x2 – 5x – 4 = 0
3. Solve the equation using any method. By Quadratic Formula, identify the values of a, b, and c. a = 1, b = – 5, c = – 4
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
=
−(−5) ± (−5)2 −4(1)(−4)
2(1)
=
5 ± 25 + 16
2
=
5 ± 41
2
𝑥 =
𝟓+ 𝟒𝟏
𝟐
𝑥 =
𝟓− 𝟒𝟏
𝟐
• The solution set of the equation is
𝟓+ 𝟒𝟏
𝟐
,
𝟓− 𝟒𝟏
𝟐
.