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1. Grade 9 – Mathematics
Quarter I
SOLVING PROBLEMS INVOLVING
QUADRATIC EQUATIONS

2. Objective:
•solve problems involving
quadratic equations.

3. Problem No. 1: A rectangular table has an area of 27 ft2 and a
perimeter of 24 ft. What are the dimensions of the table?
Understand the problem.
𝑙𝑤 = 27, represents its area.
2𝑙 + 2𝑤 = 24, represents its perimeter.
Write the equation.
𝑥2 +
𝑏
𝑎
𝑥 +
𝑐
𝑎
= 0, where −
𝑏
𝑎
is the sum of roots and
𝑐
𝑎
is the product.
−
𝑏
𝑎
= 12 or
𝑏
𝑎
= -12
𝑐
𝑎
= 27

4. Problem No. 1: (cont.) A rectangular table has an area of 27 ft2
and a perimeter of 24 ft. what are the dimension of the table?
Solve the equation.
𝑥2 +
𝑏
𝑎
𝑥 +
𝑐
𝑎
= 0 𝑥2 + (−12)𝑥 + 27 = 0
𝑥2 − 12𝑥 + 27 = 0
(𝑥 − 3)(𝑥 − 9) = 0
𝑥 − 3 = 0 𝑜𝑟 𝑥 − 9 = 0
𝑥 = 3 𝑜𝑟 𝑥 = 9
This implies that the width of the table is 3 ft. and its length is 9 ft.
−
𝑏
𝑎
= 12 or
𝑏
𝑎
= -12
𝑐
𝑎
= 27

5. Problem No. 2: The sum of two numbers is 19. The sum of the
squares of the number is 193. Find the two numbers.
Understand the problem.
𝑙𝑒𝑡 𝑥, one of the numbers.
19 − 𝑥, the other number.
Write the equation.
𝑥2
+ 19 − 𝑥 2
, the sum of the squares of the two numbers.
𝑥2
+ 19 − 𝑥 2
= 193

6. Problem No. 2: (cont.) The sum of two numbers is 19. The sum
of the squares of the number is 193. Find the two numbers
Solve the equation.
𝑥2
+ 19 − 𝑥 2
= 193
𝑥2
+ 361 − 38𝑥 + 𝑥2
= 193
2𝑥2 − 38𝑥 + 168 = 0
𝑥2 − 19𝑥 + 84 = 0
(𝑥 − 7)(𝑥 − 12) = 0
x − 7 = 0 𝑜𝑟 𝑥 − 12 = 0
x = 7 𝑜𝑟 𝑥 = 12
𝑙𝑒𝑡 𝑥, one of the numbers.
19 − 𝑥, the other number.
x = 7, 19 − 𝑥 = 19 − 7 = 12
x = 12, 19 − 𝑥 = 19 − 12 = 7
Thus, the two numbers are
7 and 12.

7. Problem No. 3: The inclined ramp used for loading cars in a
truck is 8 meters longer than the “rise of the ramp”. The “run” is
7 meters longer than the “rise”. How long are the three sides of
the ramp.
Understand the problem.
Let x = the rise of the ramp (height of a triangle, a)
x + 7 = the run of the ramp (base triangle, b)
x + 8 = the length of the ramp (hypotenuse of triangle, c).

8. Problem No. 3: (cont.) The inclined ramp used for loading cars
in a truck is 8 meters longer than the “rise of the ramp”. The
“run” is 7 meters longer than the “rise”. How long are the three
sides of the ramp.
Write the equation.
c = x + 8
b = x + 7
a = x
𝑐2
= 𝑎2
+ 𝑏2
(𝑥 + 8)2= 𝑥2 + (𝑥 + 7)2

9. Problem No. 3: (cont.) The inclined ramp used for loading cars
in a truck is 8 meters longer than the “rise of the ramp”. The
“run” is 7 meters longer than the “rise”. How long are the three
sides of the ramp.
Solve the Equation.
(𝑥 + 8)2= 𝑥2 + (𝑥 + 7)2
𝑥2
+ 16𝑥 + 64 = 𝑥2
+ 𝑥2
+ 14𝑥 + 49
0= 2𝑥2
+ 14𝑥 + 49 − 𝑥2
−16𝑥 − 64
𝑥2 − 2𝑥 − 15 = 0
𝑥2
− 2𝑥 − 15 = 0
(𝑥 − 5)(𝑥 + 3) = 0
𝑥 − 5 = 0 𝑜𝑟 𝑥 + 3 = 0
𝑥 = 5 𝑜𝑟 𝑥 = −3
Since the length of the side cannot be negative,
we conclude that the height of the ramp is 5
meters. Thus a= 5 m, b = 12 m, and c = 13 m.

10. c = x + 8
b = x + 7
a = x
Since the length of the side cannot be negative,
we conclude that the height of the ramp is 5
meters. Thus a= 5 m, b = 12 m, and c = 13 m.