# Word Problems on Quadratic Equations

1. The sum of squares of two consecutive odd natural numbers is 74. Find the numbers.

Solution: Let two consecutive odd natural numbers be x and x + 2. Since, sum of their squares is 74. we have

x2 + (x + 2)2 = 74

x2 + x2 + 4x + 4 = 74

2x2 + 4x - 70 = 0

x2 + 2x - 35 = 0

x2 + 7x - 5x - 35 = 0

x(x + 7) - 5(x + 7) = 0

(x + 7)(x - 5) = 0

Therefore, x + 7 = 0 or x - 5 = 0

x = -7 or x = 5

Now, x can not be negative as it is a natural number. Hence x = 5.

So, the numbers are 5 and 7.

2. The sum of the areas of two square fields is 468 m2. If the difference of their perimeter is 24 m, find the sides of the two squares.

Solution: Let the sides of the bigger square be x and that of the smaller square be y.

Hence, perimeter of bigger square = 4x and perimeter of smaller square = 4y.

Therefore, 4x - 4y = 24

x - y = 6

x = y + 6 ... (1)

Also, since sum of areas of two squares is 468 m2

x2 + y2 = 468 ... (2)

Substituting value of x from (1) into (2), we get

(y + 6)2 + y2 = 468

y2 + 12y + 36 + y2 = 468

2y2 + 12y - 432 = 0

y2 + 6y - 216 = 0

y = 12 or -18

Since, the side of square can not be negative, so y = 12.

Therefore,

x = y + 6 = 12 + 6 = 18

Hence, sides of squares are 18 m and 12 m.

3. The product of digits of a two digit number is 12. When 9 is added to the number, the digits interchange their places. Determine the number.

Solution: Let the digit at ten's place be x and digit at unit's place be y.

Therefore, number = 10x + y

When digits are interchanged, the number becomes 10y + x

Therefore, 10x + y + 9 = 10y + x

10x - x + y - 10y = -9

9x - 9y = -9

x - y = -1

x = y - 1 ... (1)

Also, product of digits is 12. Hence,

xy = 12 ... (2)

Substituting value of x from (1) into (2), we get

(y - 1)y = 12

y2 - y - 12 = 0

(y - 4)(y + 3) = 0

y = 4 or y = -3

Since, digit can not be negative, y = 4.

Hence x = y - 1 = 4 - 1 = 3

Therefore, the number is 34.

4. The sum of two natural numbers is 12. If sum of their reciprocals is 4/9, find the numbers.

Solution: Let one number be x.

Therefore, other number = 12 - x.

Since, sum of their reciprocals is 4/9, we get

$$\frac{1}{x} + \frac{1}{12 - x} = \frac{4}{9}$$

$$\frac{12 - x + x}{x(12 - x)} = \frac{4}{9}$$

$$\frac{12}{12x - x^2} = \frac{4}{9}$$

27 = 12x - x2

x2 - 12x + 27 = 0

(x - 3)(x - 9) = 0

x = 3 or x = 9

When first number x is 3, other number is 12 - 3 = 9 and when first number x is 9, other number is 12 - 9 = 3.

Therefore, the required numbers are 3 and 9.