In this page quadratic equation solution6 we are going to see solution of the word problems of the topic quadratic equation.

Question 11

The sum of the number and its positive square root is 7/36. Find the number.

Solution:

Let "x" be the required number

The sum of the number and its positive square root is 7/36.

x + √x = 7/36

√x = (7/36) - x

x = [(7/36) - x]²

x = (7/36)² - 2 (7/36) x + x²

x = (49/1296) - (7x/18) + x²

1296 x = 49 - 504 x + 1296 x²

49 - 504 x + 1296 x² = 1296 x

1296 x² - 504 x - 1296 x + 49 = 0

1296 x² - 1800 x + 49 = 0

1296 x² - 1764 x - 36 x + 49 = 0

36 x (36 x - 49) - 1 (36 x - 49) = 0

(36 x - 1) (36 x - 49) = 0

36 x - 1 = 0               36 x - 49 = 0

36 x = 1                      36 x = 49

x = 1/36                     x = 49/36

Therefore the required number is 1/36

Verification:

The sum of the number and its positive square root is 7/36.

(1/36) + √(1/36) = 7/36

(1/36) + (1/6) = 7/36

taking L.C.M we get

(1 + 6)/36 = 7/36

7/36 = 7/36

Question 12

The sum of two numbers is 15 and the sum of its reciprocals is 3/10. Find the numbers.

Solution:

Let "x" and "y" are the required two numbers

The sum of two numbers is 15

x + y = 15

y = 15 - x -----(1)

the sum of its reciprocals is 3/10

(1/x) + (1/y) = 3/10

(y + x)/x y = 3/10

10 (x + y) = 3 x y

10 x + 10 y = 3 x y  ---- (2)

Now we are going to apply the value of y in the second equation

10 x + 10 (15 - x) = 3 x (15 - x)

10 x + 150 - 10 x = 45 x - 3 x²

3 x² - 45 x + 150 = 0

3 x² - 15 x - 30 x + 150 = 0

3 x (x - 5) - 30 (x - 5) = 0

(3 x - 30) (x - 5) = 0

3 x - 30 = 0              x - 5 = 0

3 x = 30                     x = 5

x = 30/3

x = 10

Therefore the two numbers are 10 and 5.

Verification:

The sum of two numbers is 15

10 + 5 = 15

15 = 15

the sum of its reciprocals is 3/10

(1/10) + (1/5) = 3/10

Taking l.C.M we get,

(1 + 2)/10 = 3/10

3/10  = 3/10  