# 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

x^{2} + (x + 2)^{2} = 74

x^{2} + x^{2} + 4x + 4 = 74

2x^{2} + 4x - 70 = 0

x^{2} + 2x - 35 = 0

x^{2} + 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 m^{2}. 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 m^{2}

x^{2} + y^{2} = 468 ... (2)

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

(y + 6)^{2} + y^{2} = 468

y^{2} + 12y + 36 + y^{2} = 468

2y^{2} + 12y - 432 = 0

y^{2} + 6y - 216 = 0

Solving by quadratic formula,

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

y^{2} - 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 - x^{2}

x^{2} - 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.