This document discusses methods for solving various types of equations that are non-quadratic but can be transformed into quadratic equations. It provides examples of solving rational equations using the CRAM method, as well as examples of solving equations that involve radicals, fractions, and reciprocals by making substitutions to transform them into quadratic equations in standard form. Step-by-step solutions are shown for determining the solutions or solution sets of different types of transformable equations.

2. Solving Equations
That Are
Transformable Into
Quadratic Equations
Mark Joven A. Alam-alam, LPT

3. There are equations in
mathematics that are non-
quadratic in form but can be
written in the form of a quadratic
equation.

4. Rational Equations
Equations that contain rational
expressions
CRAM method is generally used
in solving rational equations

5. CRAM Method
C - Clear all fractions by multiplying both sides of the
equation in one variable by the least common
denominator (LCD).
R - Remove all grouping symbols, if necessary.
A - Add or subtract similar terms.
M - Multiply or divide both sides by the numerical
coefficient of the variable, leaving the variable on the
left side of the equation with a coefficient equal to 1. Then
verify or check if the values of the variable to satisfy the
given equation.

6. Example 1 : Determine the solutions of each
rational equation.
a. 1 -
18
2𝑥2 = 0
b.
1
𝑥−2
+
2
𝑥+2
= 1

7. Solution:
a. 1 -
18
2𝑥2 = 0
1 −
18
2𝑥2 = 0 (2𝑥2
)
2𝑥2 - 18 = 0
2𝑥2
= 18
2𝑥2
2
=
18
2
𝑥2
= 9
𝑥2 = 9
x = ± 3
The solution set is { -3, 3}.

8. Solution:
b.
1
𝑥−2
+
2
𝑥+2
= 1
1
𝑥−2
+
2
𝑥+2
= 1 (𝑥 − 2) (𝑥 + 2)
𝑥 + 2 + 2(𝑥 − 2) = 𝑥2
-4
𝑥 + 2 + 2𝑥 − 4 = 𝑥2
-4
3𝑥 - 2= 𝑥2
- 4
0 = 𝑥2
-3x - 2
x =
−𝑏 ± 𝑏2 −4𝑎𝑐
2𝑎
x =
−(−3) ± (−3)2 −4(1)(−2)
2(1)
x =
3 ± 9+8
2
x =
3 ± 17
2
The solution set is {
3+ 17
2
,
3 − 17
2
}.

9. There are also equations in mathematics
that are quadratic in form or reducible
to quadratic in form.
These equations are originally not
quadratic in nature but they can be
transformed into quadratic equations
using appropriate techniques.

10. Example 2 : Determine the solution set of
𝒙𝟒
- 𝟓𝒙𝟐
+ 4 = 0
Solution:
𝒙𝟒
= (𝒙𝟐
)
𝟐
Let u = 𝒙𝟐
𝑥4
-5𝑥2
+4=0 → 𝑢2
-5u+4=0
𝑢2
-5u+4=0
(u-1)(u-4) = 0
u-1 = 0
u = 1
u-4 = 0
u = 4
𝑥2
= 1
𝑥2 = 1
x = ± 1
𝑥2
= 4
𝑥2 = 4
x = ± 2
The solution set is
{ -1,1,-2,2 }.

11. Example 3 : Determine the solution set of
2𝒙
𝟐
𝟑 - 𝟓𝒙
𝟏
𝟑 + 2 = 0
Solution:
𝒙
𝟐
𝟑= (𝒙
𝟏
𝟑)
𝟐
Let u =𝒙
𝟏
𝟑
2𝒙
𝟐
𝟑-𝟓𝒙
𝟏
𝟑+2=0→ 2𝑢2
-5u+2=0
2𝑢2
-5u+2=0
(2u-1)(u-2) = 0
2u-1 = 0
2u = 1
u =
1
2
u-2 = 0
u = 2
𝒙
𝟏
𝟑=
1
2
𝒙
𝟏
𝟑=
1
2
3
x =
1
8
The solution set is {
1
8
, 8}.
𝒙
𝟏
𝟑=2
𝒙
𝟏
𝟑=2
3
x = 8

12. Example 4 : The sum of two numbers is 6 and the sum
of their reciprocals is
3
4
. Find the numbers.
Solution:
Let x = 1st number
6 – x = 2nd number
1
𝑥
= reciprocal of 1st number
1
6−𝑥
= reciprocal of 2nd number
1
𝑥
+
1
6−𝑥
=
3
4
1
𝑥
+
1
6−𝑥
=
3
4
(4x)(6-x)
4(6-x) + 4x = 3(x)(6-x)
24-4x + 4x = 18x - 3𝑥2
3𝑥2
-18x + 24 = 0

13. Example 4 : The sum of two numbers is 6 and the sum
of their reciprocals is
3
4
. Find the numbers.
Solution:
3𝑥2
-18x + 24 = 0
3𝑥2
3
-
18𝑥
3
+
24
3
= 0
𝑥2
- 6x + 8 = 0
(x-4)(x-2) = 0
𝒙 − 𝟒 = 𝟎
x = 4
𝒙 − 𝟐 = 𝟎
x = 2
If x = 4,
6-x = 6-4
= 2
If x = 2,
6-x = 6-2
= 4
Therefore, the numbers are 2
and 4.