Different types of Word Problems on Quadratic Equations by Factoring are available on this page. So, the students who feel they are lagging in the concept of a quadratic equation can make use of our page and practice the problems. Learn the unknown values of x with the help of the factorization method of a quadratic equation. Solve the real-world problems that are related to the quadratic equations from here.
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Word Problems on Quadratic Equations by Factoring
Example 1.
The school has a fund of 156. In addition, each student of the school contributes the number of rupees equal to the number of students. The total money is divided equally among the students. If each of the students gets 28 rupees. Find the number of students gets 28 find the number of students in the school.
Solution:
Let the number of students be x
Total contributions from them = x²
School has a fund of 156
According to the problem
x² + 156 = 28x
x² – 28x + 156 = 0
x² – 12x – 13x + 156 = 0
x(x – 12) – 13(x – 12) = 0
(x – 12) (x – 13) = 0
x = 12, 13.
Therefore 12 or 13 are the students in the school
The two answers are accepted in this case.
Example 2.
The product of two numbers is 6 if their sum added to the sum of their squares is 18, find the numbers.
Solution:
Let the number be x and y
As their product is 6, we get xy = 6 …….. (1)
According to the question x + y + x² + y² = 18 ……….. (2)
From equation 1; y = 6/x
y = 6/x in equation 2
x + 6/x + x² + ( 6/x)² = 18
(x + 6/x) + (x + 6/x)² – 2x. 6/x = 18
(x + 6/x)² + (x + 6/x) – 12 = 18
(x + 6/x)² + (x + 6/x) – 12 – 18 = 0
(x + 6/x)² + (x + 6/x) – 30 = 0
Putting x + 6/x = t
t² + t – 30 = 0
t² + 6t – 5t – 30 = 0
t(t + 6) -5(t + 6) = 0
t = -5, 6.
Sub -5 in x + 6/x
x + 6/x = -5
x -5x + 6 = 0
x² – 3x – 2x + 6 = 0
x(x-3) – 2(x-3) = 0
(x – 3) (x – 2) = 0
x = 3,2.
Sub 6 in x + 6/x = 0
x + 6/x = 6
x² – 6x + 6 = 0
(x – 1)(x – 6) = 0
x = 1, 6.
Therefore Solution set x = -5, 6, 1, 6.
Example 3.
Solve the quadratic equation 8m² – 8m + 2 = 0
Solution:
8m² – 8m + 2 = 0
8m² – 4m – 4m + 2 = 0
2m( 4m – 2) -1( 4m – 2) = 0
(2m – 1) (4m – 2) = 0
2m – 1 = 0 ; 4m – 2 = 0
2m = 1 ; 4m = 2
m = ½ ; m = 2/4 = ½
Therefore the solution set m = { ½, ½}
Example 4.
Solve x for equation x² + 3x – 4xy – 12y = 0
Solution:
x² + 3x – 4xy – 12y = 0
x(x + 3) – 4y(x + 3) = 0
(x – 4y) – (x + 3) = 0
x = 4y ; x = -3
Therefore the solution set x = {4y , – 3}
Example 5.
Solve the Quadratic Equation 3(x² + 1) = 7x
Solution:
3(x² + 1) = 7x
3x² + 3 = 7x
3x² + 3 – 7x = 0
3x² – 6x – x + 3 = 0
3x(x – 2) – 1(x – 2) = 0
(3x – 1) (x – 2) = 0
3x – 1 = 0 ; x – 2 = 0
3x = 1 ; x = 2
x = ⅓ ; x = 2
Therefore the solution set x = { ⅓, 2}
Example 6.
A bus travels at a certain average speed for a distance of 12km and then travels a distance of 6km at an average speed of 3 km/h more than its original speed. If it takes 2hours to complete the total journey, what is its original average speed?
Solution:
Let the original speed of train is x km/h
Time taken to cover 12 km with speed x km/h,
Time = distance Speed = 12x hours
After 12km, speed of train becomes (x + 6) km/h
Time taken to cover 6Km with speed (x + 3) km/h
Time = distance
After 63 km, speed of train becomes(x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h
Time = distance Speed = 6x + 3hours
Now as per the question
12/x + 6/x + 3 = 2
6/x + 3/x + 3 = 1
6x + 18 + 3x/x(x + 3) = 1
6x + 18 + 3x = x(x + 3)
9x + 18 = x² + 3x
x² – 9x + 3x – 18 = 0
x² – 3x – 18 = 0
x(x + 3) -6( x + 3) = 0
(x + 3) (x – 6)
x = -3, x = 6
Thus thhe solution set x = {-3, 6}
Example 7.
There are three consecutive positive integers such that the sum of the square of first and the product of the other two is 164. Find the integers.
Solution:
Let x, x+1, x + 2 are the first three consecutive integers.
The sum of the squares of the first and their product of other two is 164
Let x, (x + 1) and (x + 2) be the first three consecutive integers
The sum of the squares of first and the product of the other two is 164
x² + (x + 1)(x + 2) = 164
x² + x² + 2x + 1x + 2 = 164
2x² + 3x + 2 = 164
2x² + 3x – 162 = 0
(2x + 18)(x – 9) = 0
2x + 18 = 0
2x = 18
x = 18/2 = 9
x – 9 = 0
x = 9
Therefore the solution set is x = {18, 9}
Example 8.
Two natural numbers differ by 2 and their product is 400. Find the numbers?
Solution:
Let x and y are two natural numbers, it is differ by 2
x – y = 2
And their product is 400
So, xy = 400
y = x/400
Now apply the value of y in the first equation
x – (400/x) = 2
x² – (400/x) = 2
x² – 400 = 2x
x² – 2x – 400 = 0
(x – 20)(x + 20) = 0
x – 20 = 0
x = 20
x + 20 = 0
x = -20
Therefore the solution set x = { 20, -20}