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Solving Quadratic Equations Word Problems (Examples)

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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.

Quadratic Equations involving Numbers:

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 x^2 + x - 56 = 0 (x + 8) (x - 7) = 0 x = -8 \text{ 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 that, x^2 + (x + 5)^2 = 972x^2 + 10x + 25 - 97 = 02x^2 + 10x - 72 = 0x^2 + 5x - 36 = 0(x + 9) (x - 4) = 0x = -9\text{ 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.3x^2 = (20 - x) + 103x^2 = 30 - x3x^2 + x - 30 = 03x^2 - 9x + 10x - 30 = 03x(x - 3) + 10(x - 3) = 0(x - 3) (3x + 10) = 0x = 3, \frac{-10}{3}

Since, x cannot be equal to \frac{-10}{3} so 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 = x^2years

One year ago,

Son’s age = (x - 1) years

Man’s age = (x^2 - 1) years

It is given that one year ago; a man was 8 times as old as his son.(x^2 - 1) = 8(x - 1)x^2 - 8x - 1 + 8 = 0x^2 - 8x + 7 = 0(x - 7) (x - 1) = 0 x = 7, 1

If x = 1, then x^2 = 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 = x^2 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 = 2x^2years.

Eight years hence,

Son’s age = (x + 8) years.

Man’s age = (2x^2 +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.2x^2 + 8 = 3(x + 8) +42x^2 + 8 = 3x + 24 +42x^2 - 3x - 20 = 02x^2 - 8x + 5x - 20 = 02x(x - 4) + 5(x - 4) = 0(x - 4) (2x + 5) = 0x = 4, \frac{-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

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