In this page quadratic equation solution15 we are going to see solution of the word problems of the topic quadratic equation.

**Question 22**

The sum of the age of a father and his son is 45 years. Five years ago the product of their ages was 124. Determine their present age.

**Solution:**

Let “x” be the present age of father

Let “y” be the present age of son

The sum of the age of a father and his son is 45 years

x + y = 45

y = 45 – x ------(1)

Five years ago his father’s age = x - 5

Five years ago son’s age = y – 5

Five years ago the product of their ages was 124.

(x – 5) (y – 5) = 124

(x – 5) (45 – x – 5) = 124

(x – 5) (40 – x) = 124

40 x – x² – 200 + 5 x = 124

45 x – x² – 200 = 124

x² – 45 x + 124 + 200 = 0

x² – 45 x + 324 = 0

x² – 36 x – 9 x + 324 = 0

x (x – 36) – 9 (x – 36) = 0

(x – 9) (x – 36) = 0

x – 9 = 0 x – 36 = 0

x = 9 x = 36

Here x represent the present age of father. So x = 9 is not possible.

x = 36 is possible

To find the value of y we are going to apply the value of x in the second equation.

y = 45 – x

y = 45 - 36

y = 9

Therefore the present age of father = 36 years

Present age of son = 9 years

**Verification:**

The sum of the age of a father and his son is 45 years.

x + y = 45

36 + 9 = 45

45 = 45

Five years ago the product of their ages was 124

(36 - 5) (9 - 5) = 124

31 (4) = 124

124 = 124

quadratic equation solution15 quadratic equation solution15

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- Factoring a Quadratic Equation
- Factoring Worksheets
- Framing Quadratic Equation From Roots
- Framing Quadratic Equation Worksheet
- Remainder Theorem
- Relationship Between Coefficients and roots
- Roots of Cubic equation
- Roots of Polynomial of Degree4
- Roots of Polynomial of Degree5