# Examples of Solving Quadratic Equation Word Problems

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Solving a word problem means translating the given language into the form of a mathematical equation. We can solve the quadratic equation so obtained by the factorization method.

Let us now look at some examples of different types of real-life word problems involving quadratic equations.

#### Example 1: The product of two consecutive integers is 56. Find the integers.

Solution: Let the two consecutive integers be x and x+1.

According to the given condition we have that,

x(x + 1) = 56.

x2 + x – 56 = 0.

(x + 8) (x – 7) = 0.

x = -8 or 7.

Thus, the required integers are – 8 and -7 OR 7 and 8.

#### Example 2: Find the two natural numbers which differ by 5 and the sum of whose squares is 97.

Solution: Since the two integers differ by 5 we can assume that the two required integers are x and x+5.

According to the given condition we have,

x2 + (x + 5)2 = 97.

2x2 + 10x + 25 – 97 = 0.

2x2 + 10x – 72 = 0.

x2 + 5x – 36 = 0.

(x + 9) (x – 4) = 0.

x = -9 or 4.

Since -9 is not a natural number we conclude that x = 4. Thus, the numbers are 4 and 9.

### Quadratic Equations Involving Division into Parts:

#### Example 3: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.

Solution: Let the two parts be x and y.

According to the given conditions we have, x + y = 20 which implies that y = 10 – x.

3x2 = (20 – x) + 10.

3x2 = 30 – x.

3x2 + x – 30 = 0.

3x2 – 9x + 10x – 30 = 0.

3x(x – 3) + 10(x – 3) = 0.

(x – 3) (3x + 10) = 0.

x = 3, -10/3.

Since x cannot be equal to -10/3 we conclude that x=3.

Thus, one part is 3 and the other part is 20 – 3 = 17.

### Quadratic Equations based on Age:

#### Example 4: One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

Solution:

Let the present age of the son be x years.

So the present age of man = x2 years.

One year ago:

Son’s age = (x – 1) years.

Man’s age = (x2 – 1) years.

It is given that one year ago; a man was 8 times as old as his son.

(x2 – 1) = 8(x – 1).

x2 – 8x – 1 + 8 = 0.

x2 – 8x + 7 = 0.

(x – 7) (x – 1) = 0.

x = 7, 1.

If x=1, then x2=1, which is not possible as the father’s age cannot be equal to the son’s age. So, x = 7.

Present age of son = x years = 7 years.

Present age of man = x2 years = 49 years.

#### Example 5: The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Solution: Let the present age of the son be x years.

So the present age of man = 2x2 years.

Eight years hence:

Son’s age = (x + 8) years.

Man’s age = (2x2 +8) years.

It is given that eight years hence, the age of the father will be 4 years more than three times the age of the son.

2x2 + 8 = 3(x + 8) +4.

2x2 + 8 = 3x + 24 +4.

2x2 – 3x – 20 = 0.

2x2 – 8x + 5x – 20 = 0.

2x(x – 4) + 5(x – 4) = 0.

(x – 4) (2x + 5) = 0.

x = 4, -5/2.

But, the age cannot be negative therefore we conclude that x = 4.

Present age of son = 4 years.

Present age of father = 2(4)2 years = 32 years.