Provides example and solutions on solving problems involving quadratic equations and rational algebraic equation.

1. MATHEMATICS 9
Applications of Quadratic Equations and
Rational Algebraic Equations
Lesson 1: Problem Solving Involving Quadratic
Equations
Lesson 2: Problem Solving Involving Rational
Algebraic Equations

2. The concept of quadratic equations is illustrated in
many real – life situations. Problems that arise from
these situations, such as those involving area, work, profits, and
many others, can be solved by applying the different mathematics
concepts and principles previously studied including quadratic
equations and the different ways of solving them.
Lesson 1: Problem Solving Involving
Quadratic Equations

3. In solving problems involving quadratic equations:
1. Represent the unknown quantity or quantities using a
variable or variables.
2. Use the given facts to illustrate the problem.
3. Solve the mathematical equation.
4. Check the results.

4. Example 1: A rectangular table has an area of 27 m2 and a perimeter of 24 m.
What are the dimensions of the table?
Solution:
Let: l = length
w = width
lw = A P = 2l + 2w
lw = 27 𝟐𝟒 = 𝟐
𝟐𝟕
𝒘
+ 𝟐𝒘 → 𝟐𝟒 =
𝟓𝟒
𝒘
+ 𝟐𝒘
𝒍 =
𝟐𝟕
𝒘
𝒘 𝟐𝟒 =
𝟓𝟒
𝒘
+ 𝟐𝒘 𝒘
24w = 54 + 2w2
0 = 2w2 – 24w + 54 → w2 – 12w + 27 = 0

5. Example 1: A rectangular table has an area of 27 m2 and a
perimeter of 24 m. What are the dimensions of the
table?
Solution:
Solve for w.
w2 – 12w + 27 = 0 w – 3 = 0 w – 9 = 0
(w – 3)(w – 9) = 0 w = 3 w = 9
If we substitute w = 3 to equation 1, we have l = 9 but if we
substitute w = 9 to equation 1, we have l = 3. Therefore, the width
of the table is 3 m and its length is 9 m because the length is
longer than width.
Check:
A = lw
A = (9 m)(3 m)
A = 27 m2
P = 2l + 2w
P = 2(9 m) + 2(3 m)
P = 18 m + 6 m
P = 24 m

6. Example 2: An amusement park wants to place a new rectangular billboard to
inform visitors of their new attractions. Supposed the length of the
billboard to be placed I 4 m longer than its width and the area is 96 m2.
What will be the length and the width of the billboard?
Solution:
Let: w (width) = x
l (length) = x + 4
lw = A Solve for x.
x (x + 4) = 96 x – 8 = 0 x + 12 = 0
x2 + 4x = 96 x = 8 x = –12
x2 + 4x – 96 = 0
(x – 8)(x + 12) = 0

7. Example 2: An amusement park wants to place a new rectangular
billboard to inform visitors of their new attractions.
Supposed the length of the billboard to be placed I 4 m
longer than its width and the area is 96 m2. What will be
the length and the width of the billboard?
Solution:
The equation x2 + 4x – 96 = 0 has two solutions: x = 8 or x = –12
However, we only consider the positive value x = 8 since the
situation involves measure of length. Hence, the width of the billboard is
8 m and its length is 12 m.
w = x = 8 m l = x + 4 = 8 + 4 = 12 m
Check:
A = lw
A = (12 m)(8 m)
A = 96 m2

8. Example 3: The sum of the squares of two
consecutive integers is 85. What are
integers?
Solution:
Let: x = first integer
x + 1 = second integer
Equation: x2 + (x + 1)2 = 85
x2 + x2 + 2x + 1 = 85
2x2 + 2x + 1 – 85 = 0
2x2 + 2x – 84 = 0
x2 + x – 42 = 0
(x + 7)(x – 6) = 0
Solve for x.
x + 7 = 0 x – 6 = 0
x = –7 x = 6
The two consecutive integers
are 6 and 7 or –7 and –6.
Check:
(6)2 + (7)2 = 36 + 49 = 85
(-7)2 + (-6)2 = 49 + 36 = 85

9. There are word problems that involve rational
algebraic equations that can be solved by
applying the concept of quadratic equations. Some word
problems that involve rational algebraic equations are
distance – speed – time relationships and work problems
involving more than one person.
Lesson 2: Problem Solving Involving
Rational Algebraic Equations

10. In solving problems involving equations transformable to quadratic
equations including rational algebraic equations, follow these steps:
1. Represent the unknown quantity or quantities using a
variable or variables.
2. Use the given facts to illustrate the problem.
3. Solve the mathematical equation.
4. Check the results.

11. Example 1: A mason’s helper requires 4 hours more to pave a concrete walk than it takes
the mason. The two worked together for 3 hours, then the mason was called
away. The helper continued to work and completed the job in 2 hours. How
long would it take each to do the same job if they work separately?
Solution:
Let: x = number of hours it takes the mason to do the job
x + 4 = number of hours it takes the helper to do the job
𝟏
𝒙
= part of the job done by the mason in one hour
𝟏
𝒙+𝟒
= part of the job done by the helper in one hour
Since the mason worked for 3 hours and the helper worked
for 5 hours to complete the job, then the equation is
𝟑
𝒙
+
𝟓
𝒙 + 𝟒
= 𝟏

12. Example 1:
Solution:
LCD = x(x + 4)
𝟑
𝒙
+
𝟓
𝒙+𝟒
= 𝟏
𝒙 𝒙 + 𝟒
𝟑
𝒙
+
𝟓
𝒙+𝟒
= 𝟏(𝒙)(𝒙 + 𝟒)
3(x + 4) + 5(x) = x(x + 4)
3x + 12 + 5x = x2 + 4x
8x + 12 = x2 + 4x
0 = x2 + 4x – 8x – 12
0 = x2 – 4x – 12
0 = (x – 6)(x + 2)
Solve for x.
x – 6 = 0 x + 2 = 0
x = 6 x = –2
We only get the positive
value of x since the problem
requires the number of hours.
Therefore, the mason can do
the job alone in 6 hours,
while the helper can do the
same job in 10 hours.

13. Example 2: A man drives 500 km to a business convention. On the return trip, he
increases his speed by 25 km per hour and saves 1 hour driving time.
How fast did he go on each trip?
Solution:
Let: x = speed in going to the convention
x + 25 = speed of the return trip
𝟓𝟎𝟎
𝒙
= length of time in going to the convention
𝟓𝟎𝟎
𝒙+𝟐𝟓
= length of time in returning from the convention
Equation:
(length of time in going) = (length of time in returning) + 1
𝟓𝟎𝟎
𝒙
=
𝟓𝟎𝟎
𝒙 + 𝟐𝟓
+ 𝟏

14. Example 2:
Solution:
LCD = x(x + 25)
𝟓𝟎𝟎
𝒙
=
𝟓𝟎𝟎
𝒙+𝟐𝟓
+ 𝟏
𝒙 𝒙 + 𝟐𝟓
𝟓𝟎𝟎
𝒙
=
𝟓𝟎𝟎
𝒙+𝟐𝟓
+ 𝟏 (𝒙)(𝒙 + 𝟐𝟓)
500(x + 25) = 500(x) + 1(x)(x + 25)
500x + 12, 500 = 500x + x2 + 25x
500x + 12, 500 = x2 + 525x
0 = x2 + 525x – 500x – 12, 500
0 = x2 + 25x – 12, 500
0 = (x + 125)(x – 100)
Solve for x.
x + 125 = 0
x = –125
x – 100 = 0
x = 100
Speed is another
measurement that cannot be
negative. Therefore, the
man drives 100 kph in
going to the convention and
returns at 125 kph.